message stringlengths 2 49.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 446 108k | cluster float64 13 13 | __index_level_0__ int64 892 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 ≤ k ≤ 109).
Output
You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
from collections import defaultdict
k=int(input())
mask=0
d=defaultdict(lambda:0)
while(mask<=30):
if k&(1<<mask):
d[mask]=1
ma=mask
mask+=1
adj=defaultdict(lambda:"N")
currl=1
currvu=3
prevu=[1]
prevl=[]
currvl=4
m=4
while((currl//2)<=ma):
if d[currl//2]:
adj[currvu,currvl]="Y"
adj[currvl, currvu] = "Y"
for j in prevu:
adj[currvu, j] = "Y"
adj[j, currvu] = "Y"
for j in prevl:
adj[currvl, j] = "Y"
adj[j, currvl] = "Y"
if ((currl+2)//2)<=ma:
prevu=[currvl+1,currvl+2]
for j in prevu:
adj[currvu, j] = "Y"
adj[j, currvu] = "Y"
prevl=[currvl+3]
for j in prevl:
adj[currvl, j] = "Y"
adj[j, currvl] = "Y"
currvu=currvl+4
currvl=currvu+1
m=max(m,currvl)
else:
break
currl+=2
adj[2,currvl]="Y"
adj[currvl,2]="Y"
for i in range(1,m+1):
for j in range(1,m+1):
print(adj[i,j],end="")
print()
``` | instruction | 0 | 93,015 | 13 | 186,030 |
No | output | 1 | 93,015 | 13 | 186,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 ≤ k ≤ 109).
Output
You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
k = int(input())
edges = [['N' for i in range(1010)] for j in range(1010)]
vertices = 2
def add_edge(a, b):
global edges
edges[a][b] = edges[b][a] = 'Y'
for i in range(1, 29 + 1):
vertices += 3
add_edge(i * 3, i * 3 - 1)
add_edge(i * 3, i * 3 + 2)
add_edge(i * 3 + 1, i * 3 - 1)
add_edge(i * 3 + 1, i * 3 + 2)
for bit in range(30):
if (1 << bit) & k:
lst = 1
for i in range((29 - bit) * 2):
vertices += 1
add_edge(lst, vertices)
lst = vertices
add_edge(lst, 3 * bit + 2)
print(vertices)
print('\n'.join(map(lambda x: ''.join(x), edges[1:])))
``` | instruction | 0 | 93,016 | 13 | 186,032 |
No | output | 1 | 93,016 | 13 | 186,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 ≤ k ≤ 109).
Output
You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
k = int(input())
edges = [['N' for i in range(1010)] for j in range(1010)]
vertices = 2
def add_edge(a, b):
global edges
edges[a][b] = edges[b][a] = 'Y'
for i in range(1, 29 + 1):
vertices += 3
add_edge(i * 3, i * 3 - 1)
add_edge(i * 3, i * 3 + 2)
add_edge(i * 3 + 1, i * 3 - 1)
add_edge(i * 3 + 1, i * 3 + 2)
for bit in range(30):
if (1 << bit) & k:
lst = 1
for i in range((29 - bit) * 2):
vertices += 1
add_edge(lst, vertices)
lst = vertices
add_edge(lst, 3 * bit + 2)
print(vertices)
if 0:
for i in range(1, vertices + 1):
print(i, ':', '\n\t', end='')
for j in range(1, vertices + 1):
if edges[i][j] == 'Y':
print(j, end=' ')
print('')
else:
print('\n'.join(map(lambda x: ''.join(x[1:vertices+1]), edges[1:])))
``` | instruction | 0 | 93,017 | 13 | 186,034 |
No | output | 1 | 93,017 | 13 | 186,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel wants to write a task for a programming contest. The task is: "You are given a simple undirected graph with n vertexes. Each its edge has unit length. You should calculate the number of shortest paths between vertex 1 and vertex 2."
Same with some writers, she wants to make an example with some certain output: for example, her birthday or the number of her boyfriend. Can you help her to make a test case with answer equal exactly to k?
Input
The first line contains a single integer k (1 ≤ k ≤ 109).
Output
You should output a graph G with n vertexes (2 ≤ n ≤ 1000). There must be exactly k shortest paths between vertex 1 and vertex 2 of the graph.
The first line must contain an integer n. Then adjacency matrix G with n rows and n columns must follow. Each element of the matrix must be 'N' or 'Y'. If Gij is 'Y', then graph G has a edge connecting vertex i and vertex j. Consider the graph vertexes are numbered from 1 to n.
The graph must be undirected and simple: Gii = 'N' and Gij = Gji must hold. And there must be at least one path between vertex 1 and vertex 2. It's guaranteed that the answer exists. If there multiple correct answers, you can output any of them.
Examples
Input
2
Output
4
NNYY
NNYY
YYNN
YYNN
Input
9
Output
8
NNYYYNNN
NNNNNYYY
YNNNNYYY
YNNNNYYY
YNNNNYYY
NYYYYNNN
NYYYYNNN
NYYYYNNN
Input
1
Output
2
NY
YN
Note
In first example, there are 2 shortest paths: 1-3-2 and 1-4-2.
In second example, there are 9 shortest paths: 1-3-6-2, 1-3-7-2, 1-3-8-2, 1-4-6-2, 1-4-7-2, 1-4-8-2, 1-5-6-2, 1-5-7-2, 1-5-8-2.
Submitted Solution:
```
k = int(input())
edges = [['N' for i in range(1010)] for j in range(1010)]
vertices = 2
def add_edge(a, b):
global edges
edges[a][b] = edges[b][a] = 'Y'
for i in range(1, 29 + 1):
vertices += 3
add_edge(i * 3, i * 3 - 1)
add_edge(i * 3, i * 3 + 2)
add_edge(i * 3 + 1, i * 3 - 1)
add_edge(i * 3 + 1, i * 3 + 2)
for bit in range(30):
if (1 << bit) & k:
lst = 1
for i in range((29 - bit) * 2):
vertices += 1
add_edge(lst, vertices)
lst = vertices
add_edge(lst, 3 * bit + 2)
print(vertices)
if False:
for i in range(1, vertices + 1):
print(i, ':', '\n\t', end='')
for j in range(1, vertices + 1):
if edges[i][j] == 'Y':
print(j, end=' ')
print('')
print('\n'.join(map(lambda x: ''.join(x[1:]), edges[1:])))
``` | instruction | 0 | 93,018 | 13 | 186,036 |
No | output | 1 | 93,018 | 13 | 186,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,115 | 13 | 186,230 |
Tags: constructive algorithms, graphs
Correct Solution:
```
from collections import defaultdict,deque,Counter,OrderedDict
import sys
sys.setrecursionlimit(20000)
def main():
n,m = map(int,input().split())
adj = [[] for i in range(n+1)]
for i in range(m):
a,b = map(int,input().split())
a,b = a-1,b-1
adj[a].append(b)
adj[b].append(a)
ans = ["c"]*(n)
visited = [0]*(n)
for i in range(n):
if len(adj[i]) == n - 1:
visited[i] = 1
ans[i] = "b"
def dfs(s,c):
if visited[s]: return
visited[s],ans[s] = 1,c
for j in adj[s]:
dfs(j,c)
if "c" in ans:
st = ans.index("c")
dfs(st,"a")
if "c" in ans:
st = ans.index("c")
dfs(st,"c")
check,cnta,cntc,cntb = True,ans.count("a"),ans.count("c"),ans.count("b")
for i in range(n):
if ans[i] == "a":
check &= (len(adj[i])==cnta+cntb-1)
elif ans[i] == "c":
check &= (len(adj[i])==cntc+cntb-1)
if check:
print("Yes\n"+"".join(ans))
else:
print("No")
if __name__ == "__main__":
main()
``` | output | 1 | 93,115 | 13 | 186,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,116 | 13 | 186,232 |
Tags: constructive algorithms, graphs
Correct Solution:
```
#!/usr/bin/python3
(n, m) = tuple(map(int, input().split()))
cnt = [0]*n
edges = []
for i in range(m):
(u, v) = tuple(map(int, input().split()))
edges.append((u, v))
cnt[u-1] += 1
cnt[v-1] += 1
edges = sorted(edges)
edge_set = set(edges)
A = set()
B = set()
C = set()
for i in range(n):
#print(A)
#print(B)
#print(C)
if cnt[i] == n-1:
B.add(i+1)
else:
in_A = True
in_C = True
for a in A:
if ((a, i+1) not in edge_set and (i+1, a) not in edge_set):
in_A = False
else:
in_C = False
for c in C:
if ((c, i+1) not in edge_set and (i+1, c) not in edge_set):
in_C = False
else:
in_A = False
if in_A == True and in_C == True: #?
A.add(i+1)
elif in_A == False and in_C == True:
C.add(i+1)
elif in_A == True and in_C == False:
A.add(i+1)
else:
print("No")
break
else:
print("Yes")
ans = ""
for i in range(1, n+1):
if i in A:
ans += 'a'
if i in B:
ans += 'b'
if i in C:
ans += 'c'
print(ans)
``` | output | 1 | 93,116 | 13 | 186,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,117 | 13 | 186,234 |
Tags: constructive algorithms, graphs
Correct Solution:
```
edge=[[0 for _ in range(510)] for _ in range(510)]
cnt = [0]*510
s=[0]*510
n,m = map(int,input().split())
for _ in range(m):
u,v = map(int,input().split())
u-=1
v-=1
edge[u][v]=1
edge[v][u]=1
cnt[u]+=1
cnt[v]+=1
for i in range(n):
if(cnt[i]==n-1):s[i]='b'
for i in range(n):
if s[i]==0:
s[i]='a'
#print(i)
for j in range(n):
if s[j]==0:
#print(j)
#print(edge[i][j])
if edge[i][j]!=0:s[j]='a'
else:s[j]='c'
#print(s[j])
for i in range(n):
for j in range(n):
if i == j : continue
if (abs(ord(s[i])-ord(s[j])==2 and edge[i][j]==1))or(abs(ord(s[i])-ord(s[j]))<2 and edge[i][j]==0):print('No');exit()
print('Yes')
no = ''
for i in s:
if i == 0:break
no+=str(i)
print(no)
``` | output | 1 | 93,117 | 13 | 186,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,118 | 13 | 186,236 |
Tags: constructive algorithms, graphs
Correct Solution:
```
def dfs(v, graph, used, str_arr, color=0):
used[v] = True
str_arr[v] = color
for u in graph[v]:
if used[u] and str_arr[u] == color:
return False
if not used[u] and not dfs(u, graph, used, str_arr, color ^ 1):
return False
return True
def main():
n, m = list(map(int, input().strip().split()))
graph_adj = [[0] * n for i in range(n)]
for i in range(m):
a, b = list(map(int, input().strip().split()))
a -= 1
b -= 1
graph_adj[a][b] = 1
graph_adj[b][a] = 1
arr_adj = [[] for i in range(n)]
for i in range(n):
for j in range(i + 1, n):
if graph_adj[i][j] == 0:
arr_adj[i].append(j)
arr_adj[j].append(i)
used = [False] * n
str_arr = [-1] * n
#print(arr_adj)
for i in range(n):
if not used[i] and len(arr_adj[i]) > 0:
if not dfs(i, arr_adj, used, str_arr):
print("No")
return
#print(str_arr)
for i in range(n):
if str_arr[i] == -1:
str_arr[i] = 'b'
elif str_arr[i] == 0:
str_arr[i] = 'a'
else:
str_arr[i] = 'c'
for i in range(n):
for j in range(i + 1, n):
if graph_adj[i][j] and (str_arr[i] == 'a' and str_arr[j] == 'c' or str_arr[i] == 'c' and str_arr[j] == 'a'):
print("No")
return
print("Yes")
print(''.join(str_arr))
main()
``` | output | 1 | 93,118 | 13 | 186,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,119 | 13 | 186,238 |
Tags: constructive algorithms, graphs
Correct Solution:
```
color = [-1] * 510
vis = [0] * 510
def dfs(start):
vis[start] = 1
for i in g[start]:
if vis[i] == 0:
color[i] = 1 - color[start]
dfs(i)
if vis[i] != 0:
if (color[i] == color[start]):
print("No")
quit()
g = {}
for i in range(510):
g[i] = []
h = []
for i in range(510):
h.append([0] * 510)
cosas = []
a, b = map(int, input().split(' '))
for c in range(b):
d, e = map(int, input().split(' '))
h[d][e] = 1
h[e][d] = 1
cosas.append([d, e])
for i in range(1, a+1):
for j in range(1, a+1):
if h[i][j] != 1 and i != j:
g[i].append(j)
for i in range(1, a+1):
if (len(g[i]) == 0):
color[i] = -1
vis[i] = 1
if (vis[i] == 0):
color[i] = 0
dfs(i)
estrenga = ""
for i in range(1, a+1):
if color[i] == 0:
estrenga += 'a'
elif color[i] == 1:
estrenga += 'c'
else:
estrenga += 'b'
for i in cosas:
arra = [i[0], i[1]]
arrb = [estrenga[arra[0]-1], estrenga[arra[1]-1]]
arrb.sort()
if arrb[0] == 'a' and arrb[1] == 'c':
print("No")
quit()
print("Yes")
print(estrenga)
``` | output | 1 | 93,119 | 13 | 186,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,120 | 13 | 186,240 |
Tags: constructive algorithms, graphs
Correct Solution:
```
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2016 missingdays <missingdays@missingdays>
#
# Distributed under terms of the MIT license.
"""
"""
from string import ascii_lowercase
def indx(c):
global ascii_lowercase
return ascii_lowercase.index(c)
def read_list():
return [int(i) for i in input().split()]
def new_list(n):
return [0 for i in range(n)]
def new_matrix(n, m=0):
return [[0 for i in range(m)] for i in range(n)]
n, m = read_list()
n += 1
string = new_list(n)
v = new_matrix(n, n)
for i in range(m):
v1, v2 = read_list()
v[v1][v2] = True
v[v2][v1] = True
for i in range(1, n):
l = v[i].count(True)
if l == n-2:
string[i] = 'b'
for i in range(1, n):
if string[i]:
continue
string[i] = 'a'
for j in range(1, n):
if string[j]:
continue
if v[i][j] == True:
string[i] = 'a'
else:
string[i] = 'c'
for i in range(1, n):
for j in range(1, n):
if i == j:
continue
if (abs(indx(string[i])-indx(string[j])) == 2 and v[i][j]) or (abs(indx(string[i])-indx(string[j])) < 2 and not v[i][j]):
print("No")
exit()
print("Yes")
for i in range(1, n):
print(string[i], end="")
print()
``` | output | 1 | 93,120 | 13 | 186,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,121 | 13 | 186,242 |
Tags: constructive algorithms, graphs
Correct Solution:
```
def dfs(v):
visit[v] = cnt
for u in vertex[v]:
if not visit[u] and u in challengers:
dfs(u)
n, m = map(int, input().split())
vertex = [[] for i in range(n + 1)]
challengers = set()
ans = [''] * (n + 1)
middle = set()
for i in range(m):
a, b = map(int, input().split())
vertex[a].append(b)
vertex[b].append(a)
for i in range(1, n + 1):
s = set(j for j in range(1, n + 1))
s.discard(i)
if s == set(vertex[i]):
ans[i] = 'b'
middle.add(i)
else:
challengers.add(i)
visit = [0] * (n + 1)
cnt = 0
for c in challengers:
if not visit[c]:
cnt += 1
dfs(c)
if cnt > 2 or cnt == 1:
print('No')
elif cnt == 2:
first = set()
second = set()
for i in range(1, n + 1):
if visit[i] == 1:
first.add(i)
elif visit[i] == 2:
second.add(i)
for c in first:
s = first
s.discard(c)
if set(vertex[c]) - middle != s:
print('No')
break
s.add(c)
else:
for c in first:
ans[c] = 'a'
for c in second:
s = second
s.discard(c)
if set(vertex[c]) - middle != s:
print('No')
break
s.add(c)
else:
for c in second:
ans[c] = 'c'
if not ans[1:].count(''):
print('Yes', ''.join(ans[1:]), sep = '\n')
``` | output | 1 | 93,121 | 13 | 186,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c. | instruction | 0 | 93,122 | 13 | 186,244 |
Tags: constructive algorithms, graphs
Correct Solution:
```
import collections
n, m = map(int, input().split())
ans, A, B, C = [''] * n, [], [], []
d = collections.defaultdict(list)
for i in range(m):
x, y = map(int, input().split())
d[y - 1].append(x - 1)
d[x - 1].append(y - 1)
for k, v in d.items(): #set 'b'
if len(v) == n - 1:
ans[k] = 'b'
B.append(k)
for i in range(n): #set 'a'
if ans[i] == '':
ans[i] = 'a'
A.append(i)
de = collections.deque()
de.append(i)
while(len(de) != 0):
cur = de.popleft()
for j in d[cur]:
if ans[j] == '':
ans[j] = 'a'
A.append(j)
de.append(j)
break
for i in range(n): #set 'c'
if ans[i] == '':
ans[i] = 'c'
C.append(i)
temp = sorted(A + B)
for i in A: #check 'a'
d[i].append(i)
if sorted(d[i]) != temp:
print('No')
exit(0)
temp = sorted(B + C)
for i in C:
d[i].append(i)
if sorted(d[i]) != temp:
print('No')
exit(0)
print('Yes')
print(''.join(ans))
``` | output | 1 | 93,122 | 13 | 186,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
#!/usr/bin/python3
(n, m) = tuple(map(int, input().split()))
cnt = [0]*n
edges = []
for i in range(m):
(u, v) = tuple(map(int, input().split()))
edges.append((u, v))
cnt[u-1] += 1
cnt[v-1] += 1
edge_set = set(edges)
A = set()
B = set()
C = set()
for i in range(n):
if cnt[i] == n-1:
B.add(i+1)
else:
in_A = True
in_C = True
for a in A:
if ((a, i+1) not in edge_set and (i+1, a) not in edge_set):
in_A = False
else:
in_C = False
for c in C:
if ((c, i+1) not in edge_set and (i+1, c) not in edge_set):
in_C = False
else:
in_A = False
if in_A == True and in_C == True:
A.add(i+1)
elif in_A == False and in_C == True:
C.add(i+1)
elif in_A == True and in_C == False:
A.add(i+1)
else:
print("No")
break
else:
print("Yes")
ans = ""
for i in range(1, n+1):
if i in A:
ans += 'a'
if i in B:
ans += 'b'
if i in C:
ans += 'c'
print(ans)
``` | instruction | 0 | 93,123 | 13 | 186,246 |
Yes | output | 1 | 93,123 | 13 | 186,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
edge=[[0 for _ in range(510)] for _ in range(510)]
cnt = [0]*510
s=[0]*510
n,m = map(int,input().split())
for _ in range(m):
u,v = map(int,input().split())
u-=1;v-=1
edge[u][v]=1;edge[v][u]=1
cnt[u]+=1;cnt[v]+=1
for i in range(n):
if(cnt[i]==n-1):s[i]='b'
for i in range(n):
if s[i]==0:
s[i]='a'
for j in range(n):
if s[j]==0:
if edge[i][j]!=0:s[j]='a'
else:s[j]='c'
for i in range(n):
for j in range(n):
if i == j : continue
if (abs(ord(s[i])-ord(s[j])==2 and edge[i][j]==1))or(abs(ord(s[i])-ord(s[j]))<2 and edge[i][j]==0):print('No');exit()
print('Yes')
print(''.join(s[:s.index(0)]))
``` | instruction | 0 | 93,124 | 13 | 186,248 |
Yes | output | 1 | 93,124 | 13 | 186,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
f = lambda: map(int, input().split())
n, m = f()
p = [1 << i for i in range(n)]
for i in range(m):
x, y = f()
x -= 1
y -= 1
p[x] |= 1 << y
p[y] |= 1 << x
s = set(p)
b = (1 << n) - 1
t = {}
if b in s:
t[b] = 'b'
s.remove(b)
if len(s) == 2:
a, c = s
if a | c == b and bool(a & c) == (b in t):
t[a], t[c] = 'ac'
s.clear()
print('No' if s else 'Yes\n' + ''.join(map(t.get, p)))
# Made By Mostafa_Khaled
``` | instruction | 0 | 93,125 | 13 | 186,250 |
Yes | output | 1 | 93,125 | 13 | 186,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
#!/usr/bin/python3
class StdIO:
def read_int(self):
return int(self.read_string())
def read_ints(self, sep=None):
return [int(i) for i in self.read_strings(sep)]
def read_float(self):
return float(self.read_string())
def read_floats(self, sep=None):
return [float(i) for i in self.read_strings(sep)]
def read_string(self):
return input()
def read_strings(self, sep=None):
return self.read_string().split(sep)
io = StdIO()
def dfs(adj, visited, v, word, c, component):
visited[v] = True
word[v] = c
component.append(v)
for u in adj[v]:
if not visited[u]:
dfs(adj, visited, u, word, c, component)
def main():
n, m = io.read_ints()
adj = [list() for i in range(n)]
for i in range(m):
u, v = io.read_ints()
u -= 1
v -= 1
adj[u].append(v)
adj[v].append(u)
word = [None]*n
visited = [False]*n
# print(adj)
# bs = set()
bs = 0
for v in range(n):
if len(adj[v]) == n-1:
# bs.add(v)
visited[v] = True
word[v] = 'b'
bs += 1
comp = 0
good = True
for v in range(n):
if not visited[v]:
comp += 1
if comp > 2:
good = False
break
component = []
dfs(adj, visited, v, word, ['a', 'c'][comp-1], component)
size = len(component)
good = True
for u in component:
if len(adj[u]) != size-1 + bs:
good = False
break
if not good:
break
if not good:
print('No')
return
print('Yes')
print(''.join(word))
if __name__ == '__main__':
main()
``` | instruction | 0 | 93,126 | 13 | 186,252 |
Yes | output | 1 | 93,126 | 13 | 186,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
from collections import defaultdict,deque,Counter,OrderedDict
def main():
n,m = map(int,input().split())
adj = [[] for i in range(n+1)]
for i in range(m):
a,b = map(int,input().split())
adj[a].append(b)
adj[b].append(a)
ans = ["d"]*(n+1)
visited = [0] * (n + 1)
for i in range(1,n+1):
if len(adj[i]) == n-1:
visited[i] = 1
ans[i] = "b"
def dfs(st,ck):
if visited[st]: return
visited[st] = 1
ans[st] = ck
for i in adj[st]:
dfs(i,ck)
if "d" in ans[1:]:
st = ans.index("d",1)
dfs(st,"a")
if "d" in ans[1:]:
st = ans.index("d",1)
dfs(st,"c")
ans = ans[1:]
if "d" in ans:
print("No")
else:
print("Yes")
print("".join(ans))
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,127 | 13 | 186,254 |
No | output | 1 | 93,127 | 13 | 186,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
#!/usr/bin/python3
(n, m) = tuple(map(int, input().split()))
edges = []
for i in range(m):
(u, v) = tuple(map(int, input().split()))
edges.append((u, v))
edge_set = set(edges)
A = set()
B = set()
C = set()
for i in range(n):
nums = set(range(1, n+1))
for j in edges:
if j[0] == i+1:
nums.remove(j[1])
if j[1] == i+1:
nums.remove(j[0])
if len(nums) == 1 :
B.add(i+1)
else:
in_A = True
in_C = True
for a in A:
if ((a, i+1) not in edge_set):
in_A = False
else:
in_C = False
for c in C:
if ((c, i+1) not in edge_set):
in_C = False
else:
in_A = False
if in_A == True and in_C == True: #?
A.add(i+1)
elif in_A == False and in_C == True:
C.add(i+1)
elif in_A == True and in_C == False:
A.add(i+1)
else:
print("No")
break
else:
print("Yes")
ans = ""
for i in range(1, n+1):
if i in A:
ans += 'a'
if i in B:
ans += 'b'
if i in C:
ans += 'c'
print(ans)
``` | instruction | 0 | 93,128 | 13 | 186,256 |
No | output | 1 | 93,128 | 13 | 186,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
#!/usr/bin/python3
(n, m) = tuple(map(int, input().split()))
edges = []
for i in range(m):
(u, v) = tuple(map(int, input().split()))
edges.append((u, v))
edge_set = set(edges)
A = set()
B = set()
C = set()
for i in range(n):
nums = set(range(1, n+1))
for j in edges:
if j[0] == i+1:
nums.remove(j[1])
if j[1] == i+1:
nums.remove(j[0])
if len(nums) == 1 :
B.add(i+1)
else:
if len(A) == 0 and len(C) == 0:
A.add(i+1)
elif len(A) == 0 and len(C) != 0:
for c in C:
if ((c, i+1) not in edge_set):
A.add(i+1)
break
else:
C.add(i+1)
elif len(A) != 0 and len(C) == 0:
for a in A:
if ((a, i+1) not in edge_set):
C.add(i+1)
break
else:
A.add(i+1)
else:
in_A = True
in_C = True
for a in A:
if ((a, i+1) not in edge_set):
in_A = False
for c in C:
if ((c, i+1) not in edge_set):
in_C = False
if in_A == False and in_C == True:
C.add(i+1)
elif in_A == True and in_C == False:
A.add(i+1)
else:
print("No")
break
else:
print("Yes")
ans = ""
for i in range(1, n+1):
if i in A:
ans += 'a'
if i in B:
ans += 'b'
if i in C:
ans += 'c'
print(ans)
``` | instruction | 0 | 93,129 | 13 | 186,258 |
No | output | 1 | 93,129 | 13 | 186,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
* G has exactly n vertices, numbered from 1 to n.
* For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not.
Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.
Input
The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.
Output
In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.
Examples
Input
2 1
1 2
Output
Yes
aa
Input
4 3
1 2
1 3
1 4
Output
No
Note
In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
Submitted Solution:
```
from collections import defaultdict,deque,Counter,OrderedDict
def main():
n,m = map(int,input().split())
adj = [[] for i in range(n+1)]
for i in range(m):
a,b = map(int,input().split())
adj[a].append(b)
adj[b].append(a)
ans = ["d"]*(n+1)
visited = [0] * (n + 1)
for i in range(1,n+1):
if len(adj[i]) == n-1:
visited[i] = 1
ans[i] = "b"
st = ans.index("d")
def dfs(st,ck):
if visited[st]: return
visited[st] = 1
ans[st] = ck
for i in adj[st]:
dfs(i,ck)
dfs(st,"a")
if "d" in ans:
st = ans.index("d")
dfs(st,"c")
if "d" in ans:
print("No")
else:
print("Yes")
print("".join(ans[1:]))
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,130 | 13 | 186,260 |
No | output | 1 | 93,130 | 13 | 186,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n.
You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.
Input
The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.
The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.
The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.
The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.
Output
Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.
Examples
Input
6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6
Output
5
5
14
Input
8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8
Output
27
8
19
Note
The first example is shown on the picture below:
<image>
For the first query, all subsegments of [1; 3], except this segment itself, are suitable.
For the first query, all subsegments of [4; 6], except this segment itself, are suitable.
For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].
The second example is shown on the picture below:
<image> | instruction | 0 | 93,213 | 13 | 186,426 |
Tags: binary search, data structures, dfs and similar, dsu, graphs, two pointers
Correct Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.size = n
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
if r==self.size:
r = self.num
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
mi = lambda :map(int,input().split())
li = lambda :list(mi())
n,m = mi()
edge = [[] for i in range(n)]
for _ in range(m):
u,v = mi()
edge[u-1].append(v-1)
edge[v-1].append(u-1)
parent = [-1 for i in range(n)]
depth = [0 for i in range(n)]
cnt = [0 for i in range(n)]
res = [n for i in range(n)]
for root in range(n):
if parent[root]==-1:
stack = [root]
while stack:
v = stack[-1]
if len(edge[v])==cnt[v]:
stack.pop()
else:
nv = edge[v][cnt[v]]
cnt[v] += 1
if nv==root or parent[nv]!=-1:
if depth[nv] < depth[v]-1:
m,M = nv,nv
pos = v
while pos!=nv:
m = min(pos,m)
M = max(pos,M)
pos = parent[pos]
res[m] = min(res[m],M)
else:
depth[nv] = depth[v] + 1
parent[nv] = v
stack.append(nv)
cnt = [0 for i in range(n)]
cnt[n-1] = res[n-1] - (n-1)
for i in range(n-2,-1,-1):
res[i] = min(res[i],res[i+1])
cnt[i] = res[i] - i
for i in range(1,n):
cnt[i] += cnt[i-1]
query = [[] for i in range(n+1)]
q = int(input())
for i in range(q):
l,r = mi()
query[r].append((i,l-1))
pos = -1
ans = [0 for i in range(q)]
for r in range(1,n+1):
while pos+1<n and res[pos+1]<=r:
pos += 1
for idx,l in query[r]:
if r<=pos:
tmp = cnt[r-1]
if l:
tmp -= cnt[l-1]
elif l<=pos:
tmp = cnt[pos]
if l:
tmp -= cnt[l-1]
tmp += (r-pos-1) * r - ((r+pos)*(r-pos-1)//2)
else:
tmp = (r-l) * r - ((r-1+l)*(r-l)//2)
ans[idx] = tmp
print(*ans,sep="\n")
``` | output | 1 | 93,213 | 13 | 186,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n.
You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.
Input
The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.
The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.
The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.
The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.
Output
Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.
Examples
Input
6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6
Output
5
5
14
Input
8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8
Output
27
8
19
Note
The first example is shown on the picture below:
<image>
For the first query, all subsegments of [1; 3], except this segment itself, are suitable.
For the first query, all subsegments of [4; 6], except this segment itself, are suitable.
For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].
The second example is shown on the picture below:
<image>
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.size = n
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
if r==self.size:
r = self.num
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
mi = lambda :map(int,input().split())
li = lambda :list(mi())
n,m = mi()
edge = [[] for i in range(n)]
for _ in range(m):
u,v = mi()
edge[u-1].append(v-1)
edge[v-1].append(u-1)
if n<=100:
mini = SegmentTree([10**8 for i in range(n)],min,10**8)
maxi = SegmentTree([-1 for i in range(n)],max,-1)
parent = [-1 for i in range(n)]
depth = [0 for i in range(n)]
cnt = [0 for i in range(n)]
res = [n for i in range(n)]
for root in range(n):
if parent[root]==-1:
stack = [root]
mini.update(0,root)
maxi.update(0,root)
while stack:
v = stack[-1]
if len(edge[v])==cnt[v]:
mini.update(depth[v],10**8)
maxi.update(depth[v],-1)
stack.pop()
else:
nv = edge[v][cnt[v]]
cnt[v] += 1
if nv==root or parent[nv]!=-1:
if depth[nv] < depth[v]-1:
L,R = depth[nv],depth[v]
m,M = mini.query(L,R+1),maxi.query(L,R+1)
res[m] = M
else:
depth[nv] = depth[v] + 1
parent[nv] = v
mini.update(depth[nv],nv)
maxi.update(depth[nv],nv)
stack.append(nv)
cnt = [0 for i in range(n)]
cnt[n-1] = res[n-1] - (n-1)
for i in range(n-2,-1,-1):
res[i] = min(res[i],res[i+1])
cnt[i] = res[i] - i
for i in range(1,n):
cnt[i] += cnt[i-1]
query = [[] for i in range(n+1)]
q = int(input())
for i in range(q):
l,r = mi()
query[r].append((i,l-1))
if n<=100:
pos = -1
ans = [0 for i in range(q)]
for r in range(1,n+1):
while pos+1<n and res[pos+1]<=r:
pos += 1
for idx,l in query[r]:
if r<=pos:
tmp = cnt[r-1]
if l:
tmp -= cnt[l-1]
else:
tmp = cnt[pos]
if l:
tmp -= cnt[l-1]
tmp += (r-pos-1) * r - ((r+pos)*(r-pos-1)//2)
ans[idx] = tmp
print(*ans,sep="\n")
``` | instruction | 0 | 93,214 | 13 | 186,428 |
No | output | 1 | 93,214 | 13 | 186,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n.
You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.
Input
The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.
The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.
The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.
The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.
Output
Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.
Examples
Input
6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6
Output
5
5
14
Input
8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8
Output
27
8
19
Note
The first example is shown on the picture below:
<image>
For the first query, all subsegments of [1; 3], except this segment itself, are suitable.
For the first query, all subsegments of [4; 6], except this segment itself, are suitable.
For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].
The second example is shown on the picture below:
<image>
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.size = n
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
if r==self.size:
r = self.num
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
mi = lambda :map(int,input().split())
li = lambda :list(mi())
n,m = mi()
edge = [[] for i in range(n)]
for _ in range(m):
u,v = mi()
edge[u-1].append(v-1)
edge[v-1].append(u-1)
parent = [-1 for i in range(n)]
depth = [0 for i in range(n)]
cnt = [0 for i in range(n)]
res = [n for i in range(n)]
for root in range(n):
if parent[root]==-1:
stack = [root]
while stack:
v = stack[-1]
if len(edge[v])==cnt[v]:
stack.pop()
else:
nv = edge[v][cnt[v]]
cnt[v] += 1
if nv==root or parent[nv]!=-1:
if depth[nv] < depth[v]-1:
m,M = nv,nv
pos = v
while pos!=nv:
m = min(pos,m)
M = max(pos,M)
pos = parent[pos]
res[m] = min(res[m],M)
else:
depth[nv] = depth[v] + 1
parent[nv] = v
stack.append(nv)
cnt = [0 for i in range(n)]
cnt[n-1] = res[n-1] - (n-1)
for i in range(n-2,-1,-1):
res[i] = min(res[i],res[i+1])
cnt[i] = res[i] - i
for i in range(1,n):
cnt[i] += cnt[i-1]
query = [[] for i in range(n+1)]
q = int(input())
for i in range(q):
l,r = mi()
query[r].append((i,l-1))
pos = -1
ans = [0 for i in range(q)]
for r in range(1,n+1):
while pos+1<n and res[pos+1]<=r:
pos += 1
for idx,l in query[r]:
if r<=pos:
tmp = cnt[r-1]
if l:
tmp -= cnt[l-1]
else:
tmp = cnt[pos]
if l:
tmp -= cnt[l-1]
tmp += (r-pos-1) * r - ((r+pos)*(r-pos-1)//2)
ans[idx] = tmp
print(*ans,sep="\n")
``` | instruction | 0 | 93,215 | 13 | 186,430 |
No | output | 1 | 93,215 | 13 | 186,431 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
You and AOR Ika are preparing for a graph problem in competitive programming. Generating input cases is AOR Ika-chan's job. The input case for that problem is a directed graph of the $ N $ vertices. The vertices are numbered from $ 1 $ to $ N $. Edges may contain self-loops, but not multiple edges.
However, AOR Ika suddenly lost communication and couldn't get the original graph. All that remains is the following information:
* The number of vertices with degree $ i $ is $ a_i $. That is, the number of vertices ending in $ i $ is $ a_i $.
* The number of vertices with degree $ i $ is $ b_i $. That is, the number of vertices starting from $ i $ is $ b_i $.
Is there a directed graph that is consistent with the information left behind? Judge it, output YES if it exists, and then output one consistent directed graph. If there are more than one, any one may be output. If it does not exist, output NO.
Input constraints
$ 1 \ leq n \ leq 50 $
$ 0 \ leq a_i \ leq n $
$ 0 \ leq b_i \ leq n $
sample
Sample input 1
3
1 2 0 0
1 2 0 0
Sample output 1
YES YES
0 1 0
0 0 1
0 0 0
Sample input 2
1
Ten
Ten
Sample output 2
YES YES
0
Sample input 3
2
0 2 0
0 2 0
Sample output 3
YES YES
0 1
Ten
Sample input 4
Four
1 3 0 0 0
3 0 0 1 0
Sample output 4
YES YES
0 1 1 1
0 0 0 0
0 0 0 0
0 0 0 0
Sample input 5
1
1 1
0 0
Sample output 5
NO
input
$ n $
$ a_0 \ cdots a_n $
$ b_0 \ cdots b_n $
output
If there is a graph that meets the conditions, output it in the following format. $ e_ {ij} $ should be 1 if there is a directed edge from $ i $ to $ j $, or 0 otherwise. Be careful not to print a blank at the end of the line.
YES YES
$ e_ {11} \ e_ {12} \ cdots e_ {1n} $
$ e_ {21} \ e_ {22} \ cdots e_ {2n} $
$ \ vdots $
$ e_ {n1} \ e_ {n2} \ cdots e_ {nn} $
If not, output as follows.
NO
Example
Input
3
1 2 0 0
1 2 0 0
Output
YES
0 1 0
0 0 1
0 0 0 | instruction | 0 | 93,502 | 13 | 187,004 |
"Correct Solution:
```
from operator import itemgetter
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
cnt_a = 0
cnt_b = 0
for i, (a, b) in enumerate(zip(A, B)):
cnt_a += i*a
cnt_b += i*b
if cnt_a!=cnt_b or not sum(A)==sum(B)==N:
print("NO")
exit()
B_ = []
cumi = 0
for i, b in enumerate(B):
for j in range(b):
B_.append([cumi, i])
cumi += 1
B_.sort(key=itemgetter(1), reverse=True)
Ans = [[0]*N for _ in range(N)]
cumi = 0
for i, a in enumerate(A[1:], 1):
if len(B_) < i:
print("NO")
exit()
for k in range(a):
for j in range(i):
B_[j][1] -= 1
if B_[j][1] < 0:
print("NO")
exit()
Ans[B_[j][0]][cumi] = 1
cumi += 1
B_.sort(key=itemgetter(1), reverse=True)
print("YES")
for ans in Ans:
print(*ans)
``` | output | 1 | 93,502 | 13 | 187,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,715 | 13 | 187,430 |
Tags: dp, graphs, greedy
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
ans = []
for _ in range(int(input())):
n = int(input())
c = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
res = 0
tmp = c[-1] - 1
pos_x,pos_y = 1,c[-1]
for i in range(n-1,0,-1):
tmp += 2
pos_x,pos_y = min(a[i],b[i]),max(a[i],b[i])
res = max(res,tmp+pos_y-pos_x)
if pos_x==pos_y:
tmp = c[i-1] - 1
pos_x,pos_y = 1,c[i-1]
else:
tmp += pos_x - 1 + c[i-1] - pos_y
if c[i-1] - 1 >= tmp:
tmp = c[i-1] - 1
pos_x,pos_y = 1,c[i-1]
ans.append(res)
print(*ans,sep="\n")
``` | output | 1 | 93,715 | 13 | 187,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,716 | 13 | 187,432 |
Tags: dp, graphs, greedy
Correct Solution:
```
import os,sys
from io import BytesIO,IOBase
for _ in range(int(input())):
n = int(input())
vert = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
dp,ans = -10**10,0
for i in range(1,n):
if a[i] == b[i]:
dp = 0
else:
dp = max(dp+min(a[i],b[i])-1+vert[i-1]-max(b[i],a[i])+2,abs(b[i]-a[i]))
ans = max(ans,dp+1+vert[i])
print(ans)
``` | output | 1 | 93,716 | 13 | 187,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,717 | 13 | 187,434 |
Tags: dp, graphs, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ans = abs(a[1] - b[1])
cur = ans
for i in range(1, n):
if a[i] == b[i]:
ans = max(ans, cur)
cur = c[i] + 1
else:
pres = c[i] + 1
cur = max(cur + pres - abs(a[i] - b[i]),
abs(a[i] - b[i]) + pres)
ans = max(ans, cur)
ans = max(ans, cur)
print(ans)
``` | output | 1 | 93,717 | 13 | 187,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,718 | 13 | 187,436 |
Tags: dp, graphs, greedy
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def main():
for _ in range(int(input())):
n = int(input())
vert = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
dp,ans = -10**10,0
for i in range(1,n):
if a[i] == b[i]:
dp = 0
else:
dp = max(dp+min(a[i],b[i])-1+vert[i-1]-max(b[i],a[i])+2,abs(b[i]-a[i]))
ans = max(ans,dp+1+vert[i])
print(ans)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 93,718 | 13 | 187,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,719 | 13 | 187,438 |
Tags: dp, graphs, greedy
Correct Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
c[0]=0
s=0
m=0
z=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
for k in range(1,n):
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= max(c[k]+m,s)
else:
z=max(z,s)
s=1+c[k]
m=1
print(max(z,s))
``` | output | 1 | 93,719 | 13 | 187,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,720 | 13 | 187,440 |
Tags: dp, graphs, greedy
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
def solve():
for _ in range(ii()):
n = ii()
c = li()
a = li()
b = li()
for i in range(n):
if a[i] > b[i]:
a[i],b[i] = b[i],a[i]
ans = 0
s = b[1]-a[1]+1
for i in range(1,n-1):
ans = max(ans,s+c[i])
c1 = a[i+1]+c[i]-b[i+1]
s = max(b[i+1]-a[i+1],s+c1)
if a[i+1] == b[i+1]:
s = 0
s += 1
# print(s,ans)
ans =max(ans,s+c[-1])
print(ans)
if __name__ =="__main__":
solve()
``` | output | 1 | 93,720 | 13 | 187,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,721 | 13 | 187,442 |
Tags: dp, graphs, greedy
Correct Solution:
```
for t in range(int(input())):
n=int(input())
c=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
prev=[0 for i in range(n)]
#best=[0 for i in range(n)]
best2=0
i=1
x=b[i]-a[i]
if(x>0):
temp=x+1
prev[i]=temp
elif(x==0):
prev[i]=1
else:
temp=-x+1
prev[i]=temp
#best[i]=max(prev[i]+c[i],best[i-1])
best2=max(prev[i]+c[i],best2)
for i in range(2,n):
x=b[i]-a[i]
if(x>0):
temp=max(x+1,prev[i-1]+c[i-1]-x+1)
prev[i]=temp
elif(x==0):
prev[i]=1
else:
temp=max(-x+1,prev[i-1]+c[i-1]+x+1)
prev[i]=temp
#best[i]=max(prev[i]+c[i],best[i-1])
best2=max(prev[i]+c[i],best2)
#print(best)
#print(prev)
print(best2)
``` | output | 1 | 93,721 | 13 | 187,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity. | instruction | 0 | 93,722 | 13 | 187,444 |
Tags: dp, graphs, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
maxi=0
cnt=c[n-1]
for i in range(n-1,0,-1):
if a[i]>b[i]:
a[i],b[i]=b[i],a[i]
if a[i]==b[i]:
cnt+=1
maxi=max(maxi,cnt)
cnt=c[i-1]
else:
if i==1:
cnt+=b[i]-a[i]+1
maxi=max(maxi,cnt)
else:
cnt1=cnt
cnt1+=b[i]-a[i]+1
cnt+=a[i]+(c[i-1]-b[i]+1)
maxi=max(maxi,cnt)
maxi=max(maxi,cnt1)
if cnt<c[i-1]:
cnt=c[i-1]
print(maxi)
``` | output | 1 | 93,722 | 13 | 187,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for t in range(int(input())):
n=int(input())
c=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
curlen=0
lstcur=0
ans=0
for i in range(1,n):
curlen=abs(a[i]-b[i]) +1 + c[i]
if a[i]!=b[i]:
curlen=max(curlen, c[i] + 1 + lstcur - abs(a[i]-b[i]))
lstcur=curlen
ans=max(ans,curlen)
lstcur=curlen
print(ans)
``` | instruction | 0 | 93,723 | 13 | 187,446 |
Yes | output | 1 | 93,723 | 13 | 187,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
# template begins
#####################################
from io import BytesIO, IOBase
import sys
import math
import os
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
def mint(): return map(int, input().split())
def mfloat(): return map(float, input().split())
#####################################
# template ends
# Use the recursion snippet if heavy recursion is needed (depth>1000)
def solve():
n=int(input())
*c,=mint()
*a,=mint()
*b,=mint()
# dp?
# Can we get the longest cycle ending on the given graph?
if b[1]<a[1]:a[1],b[1]=b[1],a[1]
ans=c[1]+2+b[1]-a[1]-1
current_length=ans
for i in range(2, n):
if b[i]<a[i]:a[i],b[i]=b[i],a[i]
if b[i]!=a[i]:
current_length=max(current_length+2+c[i]-1 - (b[i]-a[i]), b[i]-a[i]+2+c[i]-1)
# print(current_length, i)
else:
current_length=2+c[i]-1
ans=max(ans, current_length)
print(ans)
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,724 | 13 | 187,448 |
Yes | output | 1 | 93,724 | 13 | 187,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for tc in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
k = [0]
for i in range(1, n):
if a[i] == b[i]:
k.append(2+c[i]-1)
else:
n1 = abs(a[i]-b[i])+2+c[i]-1 # start afresh
n2 = k[i-1]-abs(a[i]-b[i])+2+c[i]-1
k.append(max(n1, n2))
print(max(k))
``` | instruction | 0 | 93,725 | 13 | 187,450 |
Yes | output | 1 | 93,725 | 13 | 187,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
s=0
m=0
z=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
for k in range(1,n):
if k==1:
if a[k]==b[k]:
s=1+c[k]
m=1
continue
else:
m=b[k]-a[k]+1
s=c[k]+m
continue
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= max(c[k]+m,s)
else:
z=max(z,s)
s=1+c[k]
m=1
print(max(z,s))
``` | instruction | 0 | 93,726 | 13 | 187,452 |
Yes | output | 1 | 93,726 | 13 | 187,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
from itertools import accumulate as ac
mod = int(1e9)+7
#mod = 998244353
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
n = ip()
c = list(inp())
a = list(inp())
b = list(inp())
dp = [0]*n
ele = []
dp[0] = c[0]-1
for i in range(1,n):
if a[i] == b[i]:
dp[i] = c[i]+2-1
else:
minus = abs(a[i]-b[i])
dp[i] = dp[i-1]-minus+ c[i]+2 -1
ele.append(c[i]+minus+1)
dp.pop(0)
ans = -float('inf')
ans = max(ans,max(dp))
if len(ele) != 0:
ans = max(ans,max(ele))
print(ans)
``` | instruction | 0 | 93,727 | 13 | 187,454 |
No | output | 1 | 93,727 | 13 | 187,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
from math import ceil
def solve(n, c, a, b):
t = 0
r = c[-1]
for i in range(n-1, 0, -1):
if a[i] == b[i]:
r+= 1
if r > t:
t = r
r = c[i-1]
else:
if r+abs(a[i]-b[i])+1 > t:
t = r + abs(a[i]-b[i])+1
if c[i]+ abs(a[i]-b[i])+1 > t:
t = c[i]+ abs(a[i]-b[i])+1
r+= c[i-1] - abs(a[i]-b[i]) +1
print(t)
for _ in range(int(input())):
n = int(input())
c = list(map(int,input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
solve(n, c, a, b)
``` | instruction | 0 | 93,728 | 13 | 187,456 |
No | output | 1 | 93,728 | 13 | 187,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
initialSpan = 0
maxSpan = 0
for i in range(1, n):
if(b[i]!=a[i]):
initialSpan+=c[i]+1
initialSpan-=abs(b[i]-a[i])
secSpan = c[i]+abs(b[i]-a[i])+1
if(secSpan>initialSpan):
maxSpan = max(maxSpan, initialSpan)
maxSpan = max(maxSpan, secSpan)
initialSpan = secSpan
else:
initialSpan = c[i]+1
maxSpan = max(maxSpan, initialSpan)
maxSpan = max(maxSpan, initialSpan)
print(maxSpan)
``` | instruction | 0 | 93,729 | 13 | 187,458 |
No | output | 1 | 93,729 | 13 | 187,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
l=[]
s=0
m=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
print(a,b,c)
for k in range(1,n):
if k==1:
if a[k]==b[k]:
l.append(s)
s=1+c[k]
m=1
l.append(s)
continue
else:
m=b[k]-a[k]+1
s=c[k]+m
l.append(s)
continue
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= c[k]+m
print("m ",m)
l.append(s)
else:
l.append(s)
s=2+c[k]
m=1
l.append(s)
print(l)
print(max(l))
``` | instruction | 0 | 93,730 | 13 | 187,460 |
No | output | 1 | 93,730 | 13 | 187,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,913 | 13 | 187,826 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
def connected_components(n, graph):
components, visited = [], [False] * n
def dfs(start):
component, stack = [], [start]
while stack:
start = stack[-1]
if visited[start]:
stack.pop()
continue
else:
visited[start] = True
component.append(start)
for i in graph[start]:
if not visited[i]:
stack.append(i)
return component
for i in range(n):
if not visited[i]:
components.append(dfs(i))
return components
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
from collections import deque
def bfs(graph, n, alpha, used, color):
"""Breadth first search on a graph!"""
q = deque([alpha])
used[alpha] = True
z, o = 0, 0
while q:
v = q.popleft()
if color[v] == -1:
color[v] = 0
z += 1
for u in graph[v]:
if not used[u]:
used[u] = True
q.append(u)
if color[u] == color[v]:return False
if color[u] == -1:
if color[v] == 1:z+=1
else: o += 1
color[u] = (color[v] + 1) % 2
return (used, color, z, o)
def shortest_path(graph, alpha, omega):
"""Returns the shortest path between two vertices!"""
used, dist, parents = bfs(graph, alpha)
if not used[omega]:
return []
path = [omega]
v = omega
while parents[v] != -1:
path += [parents[v]]
v = parents[v]
return path[::-1]
for _ in range(int(input()) if not True else 1):
#n = int(input())
n, m = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
#e, f = map(int, input().split())
graph = [[] for __ in range(n+1)]
ml = 0
for i in range(m):
x, y = map(int, input().split())
graph[x] += [y]
graph[y] += [x]
ml = max(ml, len(graph[x]), len(graph[y]))
if ml == 0:
print(3, (n*(n-1)*(n-2))//6)
continue
if ml == 1:
x=0
for i in graph:
if len(i) == 1:
x += 1
print(2, (x//2)*(n-2))
continue
# ans will be 1 if no odd cycle else 0
cn = connected_components(n+1, graph)
bipartite = True
ans = 0
used = [0]*(n+1)
color = [-1]*(n+1)
for i in range(1, len(cn)):
j = cn[i][0]
ooff = bfs(graph, n+1, j, used, color)
if not ooff:
bipartite = False
break
used, color, z, o = ooff
ans += (z*(z-1)) // 2 + (o*(o-1))//2
if not bipartite:
print(0, 1)
continue
# ans = 1
print(1, ans)
``` | output | 1 | 93,913 | 13 | 187,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,914 | 13 | 187,828 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
import sys
import threading
threading.stack_size(67108864)
sys.setrecursionlimit(100500)
def nc2(k):
return k * (k - 1) // 2
def solve():
def dfs(u):
comp[u] = n_comp
for v in g[u]:
if color[v] == 0:
color[v] = 3 - color[u]
dfs(v)
elif color[u] == color[v]:
return True
return False
def check(g, color, n):
for v in range(1, n + 1):
for u in g[v]:
if color[u] == color[v]:
return False
return True
n, m = map(int, input().split())
g = dict()
for i in range(1, n + 1):
g[i] = set()
for i in range(m):
a, b = map(int, input().split())
g[a].add(b)
g[b].add(a)
color = [0] * (n + 1)
comp = [0] * (n + 1)
n_comp = 0
for u in range(1, n + 1):
if color[u] == 0:
n_comp += 1
color[u] = 1
if dfs(u):
print(0, 1)
return
if not check(g, color, n):
print(0, 1)
return
counts = [0] * (n_comp + 1)
counts_1 = [0] * (n_comp + 1)
counts_2 = [0] * (n_comp + 1)
for i in range(1, n + 1):
counts[comp[i]] += 1
if color[i] == 1:
counts_1[comp[i]] += 1
else:
counts_2[comp[i]] += 1
c = max(counts)
if c == 1:
print(3, n * (n - 1) * (n - 2) // 6)
elif c == 2:
print(2, m * (n - 2))
elif c >= 3:
ans = 0
for i in range(1, n_comp + 1):
if counts[i] >= 3:
ans += nc2(counts_1[i]) + nc2(counts_2[i])
print(1, ans)
thread = threading.Thread(target=solve)
thread.start()
``` | output | 1 | 93,914 | 13 | 187,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,915 | 13 | 187,830 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
# 解説AC
# 二部グラフでないグラフの性質や，パスの長さを考察する
def main():
N, M = (int(i) for i in input().split())
par = [i for i in range(N)]
size = [1 for i in range(N)]
def find(x):
if par[x] == x:
return x
else:
par[x] = find(par[x])
size[x] = size[par[x]]
return par[x]
def same(x, y):
return find(x) == find(y)
def union(x, y):
x = find(x)
y = find(y)
if x == y:
return
if size[x] < size[y]:
x, y = y, x
size[x] += size[y]
par[y] = x
def get_size(x):
return size[find(x)]
G = [[] for _ in range(N)]
for _ in range(M):
a, b = (int(i) for i in input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
union(a-1, b-1)
S = [False]*4
for i in range(N):
S[min(3, get_size(i))] = True
if S[3]:
break
t = 0
if S[3]:
t = 1
elif S[2]:
t = 2
else:
t = 3
color = [-1]*N
def dfs(s):
stack = [s]
color[s] = 0
b = 1
w = 0
while stack:
v = stack.pop()
for u in G[v]:
if color[u] != -1:
if color[u] == color[v]:
return False, b*w
continue
color[u] = color[v] ^ 1
if color[u] == 0:
b += 1
else:
w += 1
stack.append(u)
return True, b*(b-1)//2 + w*(w-1)//2
is_bipartite, _ = dfs(0)
if is_bipartite:
w = 0
if t == 3:
w = N*(N-1)*(N-2)//3//2
elif t == 2:
used = [False]*N
for i in range(N):
if not used[find(i)] and get_size(i) == 2:
w += (N-2)
used[find(i)] = True
elif t == 1:
used = [False]*N
color = [-1]*N
for i in range(N):
if not used[find(i)] and get_size(i) >= 3:
_, ways = dfs(i)
w += ways
used[find(i)] = True
print(t, w)
else:
print(0, 1)
if __name__ == '__main__':
main()
``` | output | 1 | 93,915 | 13 | 187,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,916 | 13 | 187,832 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
MOD = 998244353
#MOD = 10**9+7
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
"""
Scenarios:
0) no edges, answer is (3,N choose 3)
1) already have an odd cycle: answer is (0,1)
2) no odd cycle, but a connected component of at least size 3: answer is (1,W), where W is the number of paths of length 2
number of paths of length 2 = sum(x*(x-1)//2) where x is the degree of each vertex
3) no odd cycle, no component of size 3 or more, some isolated edges: answer is (2,M*(N-2))
How to count odd cycles
"""
def solve():
N, M = getInts()
#case 0
if not M:
print(3,N*(N-1)*(N-2)//6)
return
graph = dd(set)
for m in range(M):
A, B = getInts()
A -= 1
B -= 1
graph[A].add(B)
graph[B].add(A)
parent = [-1]*N
levels = [-1]*N
edges = set()
global flag
flag = False
@bootstrap
def dfs(node,level):
global flag
global odds
global evens
levels[node] = level
if level: odds += 1
else: evens += 1
visited.add(node)
nodes.remove(node)
for neigh in graph[node]:
if neigh in visited and neigh != parent[node] and levels[neigh] == level:
flag = True
if neigh not in visited:
parent[neigh] = node
yield dfs(neigh,level^1)
yield
ans = 0
nodes = set(list(range(N)))
while nodes:
node = next(iter(nodes))
visited = set()
global odds
odds = 0
global evens
evens = 0
dfs(node,0)
ans += odds*(odds-1)//2 + evens*(evens-1)//2
if flag:
print(0,1)
return
if ans:
print(1,ans)
return
if edges:
print(1,len(edges))
return
print(2,M*(N-2))
return
#for _ in range(getInt()):
solve()
#solve()
#print(time.time()-start_time)
``` | output | 1 | 93,916 | 13 | 187,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,917 | 13 | 187,834 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
def main():
N, M = (int(i) for i in input().split())
par = [i for i in range(N)]
size = [1 for i in range(N)]
def find(x):
if par[x] == x:
return x
else:
par[x] = find(par[x])
size[x] = size[par[x]]
return par[x]
def same(x, y):
return find(x) == find(y)
def union(x, y):
x = find(x)
y = find(y)
if x == y:
return
if size[x] < size[y]:
x, y = y, x
size[x] += size[y]
par[y] = x
def get_size(x):
return size[find(x)]
G = [[] for _ in range(N)]
for _ in range(M):
a, b = (int(i) for i in input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
union(a-1, b-1)
S = [False]*4
for i in range(N):
S[min(3, get_size(i))] = True
if S[3]:
break
t = 0
if S[3]:
t = 1
elif S[2]:
t = 2
else:
t = 3
color = [-1]*N
def dfs(s):
stack = [s]
color[s] = 0
b = 1
w = 0
while stack:
v = stack.pop()
for u in G[v]:
if color[u] != -1:
if color[u] == color[v]:
return False, b*w
continue
color[u] = color[v] ^ 1
if color[u] == 0:
b += 1
else:
w += 1
stack.append(u)
return True, b*(b-1)//2 + w*(w-1)//2
is_bipartite, _ = dfs(0)
if is_bipartite:
w = 0
if t == 3:
w = N*(N-1)*(N-2)//3//2
elif t == 2:
used = [False]*N
for i in range(N):
if not used[find(i)] and get_size(i) == 2:
w += (N-2)
used[find(i)] = True
elif t == 1:
used = [False]*N
color = [-1]*N
for i in range(N):
if not used[find(i)] and get_size(i) >= 3:
_, ways = dfs(i)
w += ways
used[find(i)] = True
print(t, w)
else:
print(0, 1)
if __name__ == '__main__':
main()
``` | output | 1 | 93,917 | 13 | 187,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,918 | 13 | 187,836 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
n, m = [int(x) for x in input().split()]
E = {i:[] for i in range(n)}
for i in range(m):
u, v = [int(x)-1 for x in input().split()]
E[v].append(u)
E[u].append(v)
def dfs():
visited = [False for i in range(n)]
colour = [0 for i in range(n)]
ans = 0
for v in range(n):
if visited[v]: continue
stack = [(v, 0)]
part = [0, 0]
while stack:
node, c = stack.pop()
if not visited[node]:
part[c] += 1
visited[node] = True
colour[node] = c
stack.extend((u,c^1) for u in E[node])
elif c != colour[node]:
return (0, 1)
ans += (part[0]*(part[0] - 1) + part[1]*(part[1] - 1)) // 2
return (1, ans)
if m == 0:
print(3, n*(n-1)*(n-2)//6)
elif max(len(E[v]) for v in E) == 1:
print(2, m*(n-2))
else:
ans = dfs()
print(ans[0], ans[1])
``` | output | 1 | 93,918 | 13 | 187,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,919 | 13 | 187,838 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
def read_data():
n, m = map(int, input().split())
Es = [[] for v in range(n)]
for e in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
Es[a].append(b)
Es[b].append(a)
return n, m, Es
def solve(n, m, Es):
if m == 0:
return 3, n * (n - 1) * (n - 2) // 6
if max(map(len, Es)) == 1:
return 2, m * (n-2)
patterns = 0
color = [0] * n
for u in range(n):
if color[u]:
continue
color[u] = 1
stack = [u]
n_color = [1, 0]
while stack:
v = stack.pop()
prev_color = color[v]
for w in Es[v]:
current_color = color[w]
if current_color == prev_color:
return 0, 1
if current_color == 0:
color[w] = - prev_color
n_color[(prev_color + 1)//2] += 1
stack.append(w)
n_even = n_color[0]
n_odd = n_color[1]
patterns += n_even * (n_even - 1) // 2 + n_odd * (n_odd - 1) // 2
return 1, patterns
if __name__ == '__main__':
n, m, Es = read_data()
print(*solve(n, m, Es))
``` | output | 1 | 93,919 | 13 | 187,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice. | instruction | 0 | 93,920 | 13 | 187,840 |
Tags: combinatorics, dfs and similar, graphs, math
Correct Solution:
```
g = [[] for _ in range(100005)]
n, m = map(int, input().split(' '))
for i in range(m):
a, b = map(int, input().split(' '))
g[a].append(b)
g[b].append(a)
colour = [0] * 100005
v = [0] * 100005
cycle = False
two = True
twos = 0
ones = 0
ans = 0
for i in range(1, n+1):
cs = 0
c1 = 0
c2 = 0
if (not colour[i]):
q = [i]
colour[i] = 1
while q:
cs += 1
top = q.pop()
if colour[top] == 1:
c1 += 1
else: c2 += 1
for j in g[top]:
if colour[j] == colour[top]:
cycle = True
elif colour[j] == 0:
colour[j] = -colour[top]
q = [j] + q
if cs > 2:
two = False
ans += ((c1*(c1-1))//2 + (c2*(c2-1))//2)
if cs == 2:
twos += 1
if cs == 1:
ones += 1
if cycle:
print(0, 1)
quit()
if m == 0:
print(3, (n*(n-1)*(n-2)//6))
quit()
if two:
print(2, twos*ones + 4 * (twos * (twos - 1) // 2))
quit()
sumx = 0
for i in range(1, n+1):
ll = len(g[i])
sumx += ll * (ll-1) // 2
print(1, ans)
``` | output | 1 | 93,920 | 13 | 187,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice.
Submitted Solution:
```
import sys
import threading
threading.stack_size(67108864)
sys.setrecursionlimit(100500)
def nc2(k):
return k * (k - 1) // 2
def solve():
def dfs(u):
comp[u] = n_comp
for v in g[u]:
if color[v] == 0:
color[v] = 3 - color[u]
dfs(v)
elif color[u] == color[v]:
return True
return False
n, m = map(int, input().split())
g = dict()
for i in range(1, n + 1):
g[i] = set()
for i in range(m):
a, b = map(int, input().split())
g[a].add(b)
g[b].add(a)
bad_graph = False
color = [0] * (n + 1)
comp = [0] * (n + 1)
n_comp = 0
for u in range(1, n + 1):
if color[u] == 0:
n_comp += 1
color[u] = 1
if dfs(u):
print(0, 1)
return
counts = [0] * (n_comp + 1)
counts_1 = [0] * (n_comp + 1)
counts_2 = [0] * (n_comp + 1)
for i in range(1, n + 1):
counts[comp[i]] += 1
if color[i] == 1:
counts_1[comp[i]] += 1
else:
counts_2[comp[i]] += 1
c = max(counts)
if c == 1:
print(3, n * (n - 1) * (n - 2) // 6)
elif c == 2:
print(2, m * (n - 2))
elif c >= 3:
ans = 0
for i in range(1, n_comp + 1):
if counts[i] >= 3:
ans += nc2(counts_1[i]) + nc2(counts_2[i])
print(1, ans)
thread = threading.Thread(target=solve)
thread.start()
``` | instruction | 0 | 93,921 | 13 | 187,842 |
No | output | 1 | 93,921 | 13 | 187,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m (<image> — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn't contain any loops and parallel edges. The graph isn't necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
Input
4 4
1 2
1 3
4 2
4 3
Output
1 2
Input
3 3
1 2
2 3
3 1
Output
0 1
Input
3 0
Output
3 1
Note
The simple cycle is a cycle that doesn't contain any vertex twice.
Submitted Solution:
```
__author__ = 'Andrey'
import sys
sys.setrecursionlimit(100500)
n, m = map(int, input().split())
g = dict()
for i in range(1, n + 1):
g[i] = set()
for i in range(m):
a, b = map(int, input().split())
g[a].add(b)
g[b].add(a)
bad_graph = False
color = [0] * (n + 1)
def dfs(u):
global bad_graph
for v in g[u]:
if color[v] == 0:
color[v] = 3 - color[u]
dfs(v)
elif color[u] == color[v]:
bad_graph = True
for u in range(1, n + 1):
if color[u] == 0:
color[u] = 1
dfs(u)
if bad_graph:
print(0, 1)
else:
nothing = 0
ans = -1
total = 0
set_1 = set()
set_2 = set()
for i in range(1, n + 1):
c = 0
c_1 = 0
if color[i] == 1:
for u in g[i]:
c += 1
if u in set_1:
c_1 += 1
set_1.add(u)
if color[i] == 2:
for u in g[i]:
c += 1
if u in set_2:
c_1 += 1
set_2.add(u)
if c == 0:
nothing += 1
if ans == 1:
if c >= 2:
total += c * (c - 1) // 2 - c_1 * (c_1 - 1) // 2
if ans == 2:
if c >= 2:
total = c * (c - 1) // 2 - c_1 * (c_1 - 1) // 2
ans = 1
if c == 1 and len(g[i]) == 1:
total += n - 2
if ans == -1:
if c >= 2:
ans = 1
total = c * (c - 1) // 2 - c_1 * (c_1 - 1) // 2
if c == 1 and len(g[i]) == 1:
ans = 2
total = n - 2
if nothing == n:
print(n * (n - 1) * (n - 2) // 6)
else:
print(ans, total)
``` | instruction | 0 | 93,922 | 13 | 187,844 |
No | output | 1 | 93,922 | 13 | 187,845 |
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