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 |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4 | instruction | 0 | 55,413 | 13 | 110,826 |
"Correct Solution:
```
def bfs(dp, arr, i, visited):
q = []
q.append(i)
visited[i] = True
while q:
i = q.pop(0)
visited[i] = True
for j in arr[i]:
if dp[j] == -1:
dp[j] = dp[i] + 1
elif dp[i] == dp[j]:
return -1
if visited[j] == False:
visited[j] = True
q.append(j)
return 0
n=int(input())
i=0
arr=[]
degree = [0]*n
while i<n:
s=input()
a=[]
for j,e in enumerate(s):
if e=='1':
a.append(j)
degree[i]+=1
arr.append(a)
i+=1
res = -1
ans = -1
for i in range(n):
dp = [-1]*n
visited=[False for x in range(n)]
dp[i] = 1
res=bfs(dp, arr, i, visited)
if res == 0:
ans = max(ans , max(dp))
print(ans)
``` | output | 1 | 55,413 | 13 | 110,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
from collections import deque
n = int(input())
s = [[int(c) for c in input().strip()] for _ in range(n)]
def bfs(s, start):
plan = deque([(start, 0)])
lengths = {}
while len(plan) > 0:
vertex, length = plan.popleft()
if vertex in lengths:
if (lengths[vertex] - length) % 2 != 0:
return -1
else:
lengths[vertex] = length
plan.extend([(i, length + 1) for i, c in enumerate(s[vertex]) if c == 1 and i not in lengths])
return max(lengths.values())
bfs_result = 0
for start in range(n):
res = bfs(s, start)
if res < 0:
bfs_result = -2
break
elif res > bfs_result:
bfs_result = res
print(bfs_result + 1)
``` | instruction | 0 | 55,414 | 13 | 110,828 |
Yes | output | 1 | 55,414 | 13 | 110,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
from collections import deque
N = int(input())
S = [list(map(int, input().strip())) for _ in range(N)]
G = [[] for _ in range(N)]
for i in range(N):
for j in range(N):
if S[i][j] == 1:
G[i].append(j)
#Bipartie Graph
def is_bibarate(G):
vst = [False] * N
que = deque([0])
color = [0] * N
color[0] = 1
while len(que):
p = que.popleft()
for q in list(G[p]):
if color[q] == 0:
color[q] = - color[p]
que.append(q)
elif color[q] == color[p]:
return False
return True
#Warshal Floyd
d = [[float('inf')] * N for _ in range(N)]
for i in range(N):
for j in range(N):
if i == j:
d[i][j] = 0
elif S[i][j] == 1:
d[i][j] = 1
for k in range(N):
for i in range(N):
for j in range(N):
d[i][j] = min(d[i][j],d[i][k] + d[k][j])
wf_ans = max([max(r) for r in d]) + 1
if is_bibarate(G):
print(wf_ans)
else:
print(-1)
``` | instruction | 0 | 55,415 | 13 | 110,830 |
Yes | output | 1 | 55,415 | 13 | 110,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
'''
二部グラフ判定
'''
INFTY = 10**13
def floyd():
for k in range(N):
for i in range(N):
if d[i][k] == INFTY:
continue
for j in range(N):
if d[k][j] == INFTY:
continue
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
def dfs(v, c):
color[v] = c
for i in G[v]:
if color[i] == c:
return False
if color[i] == 0 and not dfs(i, -c):
return False
# ds[v] += d
return True
def solve():
for i in range(N):
if color[i] == 0:
if not dfs(i, 1):
return False
return True
N = int(input())
G = [[]for i in range(N)]
color = [0 for i in range(N)]
# ds = [1 for i in range(N)]
d = [[0 if i == j else INFTY for j in range(N)]for i in range(N)]
for i in range(N):
S = input()
for j, s in enumerate(S):
if s == '1':
G[i].append(j)
d[i][j] = 1
ans = solve()
if ans:
floyd()
print(max([max(di)for di in d])+1)
# print(max(ds))
else:
print(-1)
``` | instruction | 0 | 55,416 | 13 | 110,832 |
Yes | output | 1 | 55,416 | 13 | 110,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
n = int(input())
connect_matrix = [ list(input()) for i in range(n) ]
def bfs(x):
node_set = [ None for i in range(n)]
color_list = [ "White" for i in range(n) ]
search_list = [x]
d = 0
node_set[x] = 0
color_list[x] = "Black"
ans = 0
while search_list != []:
d += 1
new_search_list = []
for i in search_list:
for j, v in enumerate(connect_matrix[i]):
if v == "1":
if color_list[j] == "White":
if node_set[j] is None:
node_set[j] = node_set[i] + 1
new_search_list.append(j)
color_list[j] = "Black"
else:
if (node_set[j] + node_set[i]) % 2 == 0:
ans = -1
search_list = new_search_list
return ans, d-1
judge_list = []
dist_list = []
for i in range(n):
a = bfs(i)
judge_list.append(a[0])
dist_list.append(a[1])
if -1 in judge_list:
print(-1)
else:
print(max(dist_list)+1)
``` | instruction | 0 | 55,417 | 13 | 110,834 |
Yes | output | 1 | 55,417 | 13 | 110,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
#!/usr/bin/env python3
import sys
INF = float("inf")
from collections import deque
def solve(N: int, S: "List[List[int]]"):
visited = [False]*N
q = deque()
q.append(0)
while len(q) > 0:
node = q.popleft()
visited[node] = True
for to_, ex in enumerate(S[node]):
if ex == 0:
continue
if visited[to_] is False:
q.append(to_)
depth = [INF]*N
q = deque()
q.append((node, 1))
while len(q) > 0:
node, d = q.popleft()
depth[node] = d
# print(node, S[node])
for to_, ex in enumerate(S[node]):
if ex == 0:
continue
if depth[to_] == INF:
q.append((to_, d+1))
elif depth[to_] == d+1 or depth[to_] == d-1:
continue
else:
# print(depth)
print(-1)
return
# print(q)
print(max(depth))
return
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
S = [[int(c) for c in next(tokens)]
for _ in range(N)] # type: "List[List[int]]"
solve(N, S)
if __name__ == '__main__':
main()
``` | instruction | 0 | 55,418 | 13 | 110,836 |
No | output | 1 | 55,418 | 13 | 110,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
import sys
#input = sys.stdin.buffer.readline
from collections import deque
def main():
N = int(input())
edge = [list(str(input())) for _ in range(N)]
dist = [-1 for _ in range(N)]
dist[0] = 0
q = deque([0])
D,e = 0,0
while q:
now = q.popleft()
for i in range(N):
if edge[now][i] == "1":
fol = i
if dist[fol] == -1:
dist[fol] = dist[now]+1
if dist[fol] > D:
e = fol
dist[fol] %= 2
q.append(fol)
elif dist[now] == dist[fol]:
print(-1)
exit()
q = deque([e])
dist = [-1 for _ in range(N)]
dist[e] = 0
while q:
now = q.pop()
for i in range(N):
if edge[now][i] == "1":
fol = i
if dist[fol] == -1:
dist[fol] = dist[now]+1
q.append(fol)
print(max(dist)+1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 55,419 | 13 | 110,838 |
No | output | 1 | 55,419 | 13 | 110,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
import collections
import sys
sys.setrecursionlimit(10**8)
n=int(input())
arr=[input() for _ in range(n)]
g=[[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i==j:
continue
if arr[i][j]=='1':
g[i].append(j)
colors=[0]*n
def dfs(v,color):
colors[v]=color
for u in g[v]:
if colors[u]==color:
return False
if colors[u]==0 and not dfs(u,-color):
return False
return True
def is_bipartite():
return dfs(0,1)
if is_bipartite()==False:
print(-1)
exit()
ans=0
for v in range(n):
q=collections.deque()
q.append((v,1))
checked=[0]*n
while len(q)!=0:
tmp,depth=q.popleft()
ans=max(ans,depth)
checked[tmp]=1
for v in g[tmp]:
if checked[v]==0:
q.append((v,depth+1))
print(ans)
``` | instruction | 0 | 55,420 | 13 | 110,840 |
No | output | 1 | 55,420 | 13 | 110,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.
Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.
* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.
Constraints
* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.
Input
Input is given from Standard Input in the following format:
N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}
Output
If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.
Examples
Input
2
01
10
Output
2
Input
3
011
101
110
Output
-1
Input
6
010110
101001
010100
101000
100000
010000
Output
4
Submitted Solution:
```
n = int(input())
s = []
for _ in range(n):
s.append(input())
check = []
testc = 1000
for i in range(n):
test = 0
for c in s[i]:
test += int(c)
if test == testc:
check.append(i)
elif test < testc:
check.clear()
testc = test
check.append(i)
res = -1
if testc > 1:
check = check[0]
for i in check:
tres = 1
flag = [-2 for i in range(n)]
flag[i] = 1
que = [i]
while que:
target = que.pop()
for j in range(n):
if s[target][j] == "1":
if flag[j] == -2:
flag[j] = flag[target] + 1
que.append(j)
elif flag[j] != flag[target] + 1 and flag[j] != flag[target] - 1:
tres = -1
break
if tres == -1:
break
if tres == -1:
continue
else:
res = max(res, max(flag))
print(res)
``` | instruction | 0 | 55,421 | 13 | 110,842 |
No | output | 1 | 55,421 | 13 | 110,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today is tuesday, that means there is a dispute in JOHNNY SOLVING team again: they try to understand who is Johnny and who is Solving. That's why guys asked Umnik to help them. Umnik gave guys a connected graph with n vertices without loops and multiedges, such that a degree of any vertex is at least 3, and also he gave a number 1 ≤ k ≤ n. Because Johnny is not too smart, he promised to find a simple path with length at least n/k in the graph. In reply, Solving promised to find k simple by vertices cycles with representatives, such that:
* Length of each cycle is at least 3.
* Length of each cycle is not divisible by 3.
* In each cycle must be a representative - vertex, which belongs only to this cycle among all printed cycles.
You need to help guys resolve the dispute, for that you need to find a solution for Johnny: a simple path with length at least n/k (n is not necessarily divided by k), or solution for Solving: k cycles that satisfy all the conditions above. If there is no any solution - print -1.
Input
The first line contains three integers n, m and k (1 ≤ k ≤ n ≤ 2.5 ⋅ 10^5, 1 ≤ m ≤ 5 ⋅ 10^5)
Next m lines describe edges of the graph in format v, u (1 ≤ v, u ≤ n). It's guaranteed that v ≠ u and all m pairs are distinct.
It's guaranteed that a degree of each vertex is at least 3.
Output
Print PATH in the first line, if you solve problem for Johnny. In the second line print the number of vertices in the path c (c ≥ n/k). And in the third line print vertices describing the path in route order.
Print CYCLES in the first line, if you solve problem for Solving. In the following lines describe exactly k cycles in the following format: in the first line print the size of the cycle c (c ≥ 3). In the second line print the cycle in route order. Also, the first vertex in the cycle must be a representative.
Print -1 if there is no any solution. The total amount of printed numbers in the output must be at most 10^6. It's guaranteed, that if exists any solution then there is a correct output satisfies this restriction.
Examples
Input
4 6 2
1 2
1 3
1 4
2 3
2 4
3 4
Output
PATH
4
1 2 3 4
Input
10 18 2
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
2 3
3 4
2 4
5 6
6 7
5 7
8 9
9 10
8 10
Output
CYCLES
4
4 1 2 3
4
7 1 5 6 | instruction | 0 | 55,647 | 13 | 111,294 |
Tags: constructive algorithms, dfs and similar, graphs, math
Correct Solution:
```
import sys
s = sys.stdin.read()
inp = []
numb = 0
for i in range(len(s)):
if s[i]>='0':
numb = 10*numb + ord(s[i])-48
else:
inp.append(numb)
numb = 0
if s[-1]>='0':
inp.append(numb)
ii = 0
n = inp[ii]
ii+=1
m = inp[ii]
ii+=1
k = inp[ii]
ii+=1
coupl = [[] for _ in range(n)]
for _ in range(m):
u = inp[ii]-1
ii += 1
v = inp[ii]-1
ii += 1
coupl[u].append(v)
coupl[v].append(u)
P = [-1]*n
D = [1]*n
found = [False]*n
cycle_nodes = []
Q = [0]
while Q:
node = Q.pop()
if found[node]:
continue
found[node] = True
found_any = False
for nei in coupl[node]:
if not found[nei]:
P[nei] = node
D[nei] = D[node]+1
Q.append(nei)
found_any = True
if not found_any:
cycle_nodes.append(node)
i = max(range(n),key=lambda i:D[i])
if k*D[i]>=n:
print('PATH')
print(D[i])
out = []
while i!=-1:
out.append(i)
i = P[i]
print(' '.join(str(x+1) for x in out))
elif len(cycle_nodes)>=k:
print('CYCLES')
out = []
for i in cycle_nodes[:k]:
minD = min(D[nei] for nei in coupl[i] if (D[i] - D[nei] + 1)%3!=0)
tmp = []
if minD == D[i]-1:
a,b = [nei for nei in coupl[i] if (D[i] - D[nei] + 1)%3==0][:2]
if D[a]<D[b]:
a,b = b,a
tmp.append(i)
while a!=b:
tmp.append(a)
a = P[a]
tmp.append(a)
else:
while D[i]!=minD:
tmp.append(i)
i = P[i]
tmp.append(i)
out.append(str(len(tmp)))
out.append(' '.join(str(x+1) for x in tmp))
print('\n'.join(out))
else:
print(-1)
``` | output | 1 | 55,647 | 13 | 111,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is tuesday, that means there is a dispute in JOHNNY SOLVING team again: they try to understand who is Johnny and who is Solving. That's why guys asked Umnik to help them. Umnik gave guys a connected graph with n vertices without loops and multiedges, such that a degree of any vertex is at least 3, and also he gave a number 1 ≤ k ≤ n. Because Johnny is not too smart, he promised to find a simple path with length at least n/k in the graph. In reply, Solving promised to find k simple by vertices cycles with representatives, such that:
* Length of each cycle is at least 3.
* Length of each cycle is not divisible by 3.
* In each cycle must be a representative - vertex, which belongs only to this cycle among all printed cycles.
You need to help guys resolve the dispute, for that you need to find a solution for Johnny: a simple path with length at least n/k (n is not necessarily divided by k), or solution for Solving: k cycles that satisfy all the conditions above. If there is no any solution - print -1.
Input
The first line contains three integers n, m and k (1 ≤ k ≤ n ≤ 2.5 ⋅ 10^5, 1 ≤ m ≤ 5 ⋅ 10^5)
Next m lines describe edges of the graph in format v, u (1 ≤ v, u ≤ n). It's guaranteed that v ≠ u and all m pairs are distinct.
It's guaranteed that a degree of each vertex is at least 3.
Output
Print PATH in the first line, if you solve problem for Johnny. In the second line print the number of vertices in the path c (c ≥ n/k). And in the third line print vertices describing the path in route order.
Print CYCLES in the first line, if you solve problem for Solving. In the following lines describe exactly k cycles in the following format: in the first line print the size of the cycle c (c ≥ 3). In the second line print the cycle in route order. Also, the first vertex in the cycle must be a representative.
Print -1 if there is no any solution. The total amount of printed numbers in the output must be at most 10^6. It's guaranteed, that if exists any solution then there is a correct output satisfies this restriction.
Examples
Input
4 6 2
1 2
1 3
1 4
2 3
2 4
3 4
Output
PATH
4
1 2 3 4
Input
10 18 2
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
2 3
3 4
2 4
5 6
6 7
5 7
8 9
9 10
8 10
Output
CYCLES
4
4 1 2 3
4
7 1 5 6
Submitted Solution:
```
[n, m, k] = [int(t) for t in input().split(' ')]
e = [[] for _ in range(n)]
for _ in range(m):
[v1, v2] = [int(v) - 1 for v in input().split(' ')]
e[v1].append(v2)
e[v2].append(v1)
from_ = [None] * n
depths = [0] * n
leaves = []
def path_found(v):
print('PATH\n', depths[v])
while v != 0:
print(v+1, end=' ')
v = from_[v]
print()
exit()
def print_path(v, up):
while v != up:
print(v+1, end=' ')
v = from_[v]
print(up+1)
def print_cycle(v):
[v1, v2] = [nv for nv in e[v] if nv != from_[v]]
len1 = depths[v] - depths[v1]
len2 = depths[v] - depths[v2]
if len1 % 3 != 0:
print(len1)
return print_path(v, v1)
if len2 % 3 != 0:
print(len2)
return print_path(v, v2)
print(max(len1, len2) - min(len1, len2) + 2)
print(v+1, end=' ')
print_path(v1, v2)
def dfs(v, depth):
depths[v] = depth
if depth >= n // k:
return path_found(v)
is_leaf = True
for nv in e[v]:
if from_[nv] is None:
is_leaf = False
from_[nv] = v
dfs(nv, depth + 1)
if is_leaf:
leaves.append(v)
from_[0] = 0
dfs(0, 1)
if len(leaves) < k:
print(-1)
else:
print('CYCLES')
for i in range(k):
print_cycle(leaves[i])
``` | instruction | 0 | 55,648 | 13 | 111,296 |
No | output | 1 | 55,648 | 13 | 111,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Today is tuesday, that means there is a dispute in JOHNNY SOLVING team again: they try to understand who is Johnny and who is Solving. That's why guys asked Umnik to help them. Umnik gave guys a connected graph with n vertices without loops and multiedges, such that a degree of any vertex is at least 3, and also he gave a number 1 ≤ k ≤ n. Because Johnny is not too smart, he promised to find a simple path with length at least n/k in the graph. In reply, Solving promised to find k simple by vertices cycles with representatives, such that:
* Length of each cycle is at least 3.
* Length of each cycle is not divisible by 3.
* In each cycle must be a representative - vertex, which belongs only to this cycle among all printed cycles.
You need to help guys resolve the dispute, for that you need to find a solution for Johnny: a simple path with length at least n/k (n is not necessarily divided by k), or solution for Solving: k cycles that satisfy all the conditions above. If there is no any solution - print -1.
Input
The first line contains three integers n, m and k (1 ≤ k ≤ n ≤ 2.5 ⋅ 10^5, 1 ≤ m ≤ 5 ⋅ 10^5)
Next m lines describe edges of the graph in format v, u (1 ≤ v, u ≤ n). It's guaranteed that v ≠ u and all m pairs are distinct.
It's guaranteed that a degree of each vertex is at least 3.
Output
Print PATH in the first line, if you solve problem for Johnny. In the second line print the number of vertices in the path c (c ≥ n/k). And in the third line print vertices describing the path in route order.
Print CYCLES in the first line, if you solve problem for Solving. In the following lines describe exactly k cycles in the following format: in the first line print the size of the cycle c (c ≥ 3). In the second line print the cycle in route order. Also, the first vertex in the cycle must be a representative.
Print -1 if there is no any solution. The total amount of printed numbers in the output must be at most 10^6. It's guaranteed, that if exists any solution then there is a correct output satisfies this restriction.
Examples
Input
4 6 2
1 2
1 3
1 4
2 3
2 4
3 4
Output
PATH
4
1 2 3 4
Input
10 18 2
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
2 3
3 4
2 4
5 6
6 7
5 7
8 9
9 10
8 10
Output
CYCLES
4
4 1 2 3
4
7 1 5 6
Submitted Solution:
```
[n, m, k] = [int(t) for t in input().split(' ')]
e = [None] * n
for _ in range(m):
[v1, v2] = [int(v) - 1 for v in input().split(' ')]
if e[v1] is None: e[v1] = []
if e[v2] is None: e[v2] = []
e[v1].append(v2)
e[v2].append(v1)
from_ = [None] * n
leaves = []
def tear_up(v):
path = []
while True:
path.append(str(v+1))
if v == from_[v]: break
v = from_[v]
return path
def path_found(v, depth):
print('PATH')
path = tear_up(v)
print(len(path))
print(' '.join(path))
exit()
def get_length(v, end):
t = 1
while v != end:
v = from_[v]
t += 1
return t
def print_path(v, up):
while v != up:
print(v+1, end=' ')
v = from_[v]
print(up+1)
def print_cycle(v):
nv = [nv for nv in e[v] if nv != from_[v]]
len1 = get_length(v, nv[0])
if len1 % 3 != 0:
return print_path(v, nv[0])
len2 = get_length(v, nv[1])
if len2 % 3 != 0:
return print_path(v, nv[1])
print(v+1, end=' ')
print_path(nv[0], nv[1])
def dfs(v, depth):
if depth >= n // k:
return path_found(v, depth)
is_leaf = True
for nv in e[v]:
if from_[nv] is None:
is_leaf = False
from_[nv] = v
dfs(nv, depth + 1)
if is_leaf:
leaves.append(v)
from_[0] = 0
dfs(0, 1)
if len(leaves) < k:
print(-1)
else:
print('CYCLES')
for i in range(k):
print_cycle(leaves[i])
``` | instruction | 0 | 55,649 | 13 | 111,298 |
No | output | 1 | 55,649 | 13 | 111,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer n. Let's build a graph on vertices 1, 2, ..., n in such a way that there is an edge between vertices u and v if and only if <image>. Let d(u, v) be the shortest distance between u and v, or 0 if there is no path between them. Compute the sum of values d(u, v) over all 1 ≤ u < v ≤ n.
The gcd (greatest common divisor) of two positive integers is the maximum positive integer that divides both of the integers.
Input
Single integer n (1 ≤ n ≤ 107).
Output
Print the sum of d(u, v) over all 1 ≤ u < v ≤ n.
Examples
Input
6
Output
8
Input
10
Output
44
Note
All shortest paths in the first example:
* <image>
* <image>
* <image>
* <image>
* <image>
* <image>
There are no paths between other pairs of vertices.
The total distance is 2 + 1 + 1 + 2 + 1 + 1 = 8.
Submitted Solution:
```
n=int(input())
s = 0
b = [1] * n
c = [0] * n
c[0] = 1
bad=set()
nu = 0
for i in range(1,n):
if c[i] == 0:
j = i
while j < n:
if c[j] == 0:
c[j] = i+1
j += i+1
if c[i] == i+1:
if (i+1)*2 > n:
bad.add(i+1)
j = i
st = 1
stt = 1
while j < n:
b[j] *= i
if st % (i+1) == 0:
stt *= i+1
b[j] *= stt
j += i+1
st += 1
sdt=0
ed = (n*(n-1)) // 2
for i in range(1,n):
sdt += b[i]
ed -= sdt
sdv = 0
for i in range(2,n+1):
for j in range(i+1,n+1):
if c[i-1]*c[j-1] <= n:
sdv += 1
else:
if max(c[i-1],c[j-1])*2 > n:
nu += 1
sdv = sdv - ed
st = sdt - (n-1) - nu - sdv
s = ed + sdv*2 + st*3
print(s)
``` | instruction | 0 | 56,125 | 13 | 112,250 |
No | output | 1 | 56,125 | 13 | 112,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer n. Let's build a graph on vertices 1, 2, ..., n in such a way that there is an edge between vertices u and v if and only if <image>. Let d(u, v) be the shortest distance between u and v, or 0 if there is no path between them. Compute the sum of values d(u, v) over all 1 ≤ u < v ≤ n.
The gcd (greatest common divisor) of two positive integers is the maximum positive integer that divides both of the integers.
Input
Single integer n (1 ≤ n ≤ 107).
Output
Print the sum of d(u, v) over all 1 ≤ u < v ≤ n.
Examples
Input
6
Output
8
Input
10
Output
44
Note
All shortest paths in the first example:
* <image>
* <image>
* <image>
* <image>
* <image>
* <image>
There are no paths between other pairs of vertices.
The total distance is 2 + 1 + 1 + 2 + 1 + 1 = 8.
Submitted Solution:
```
from fractions import gcd
#from pythonds.basic import Queue
arr = list(range(1, int(input())+1))
#print(arr)
d = {}
q = []
for i in arr:
for j in arr:
if i != j and gcd(i, j) > 1:
if i in d:
d[i].append(j)
else:
d[i] = [j]
#print(d)
sum = 0
for i in d.keys():
a = [i]
q.append([i, 0])
while len(q) > 0:
[num, length] = q.pop()
for l in d[num]:
if l not in a:
sum += length+1
q.append([l, length+1])
a.append(l)
print(int(sum/2))
``` | instruction | 0 | 56,126 | 13 | 112,252 |
No | output | 1 | 56,126 | 13 | 112,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer n. Let's build a graph on vertices 1, 2, ..., n in such a way that there is an edge between vertices u and v if and only if <image>. Let d(u, v) be the shortest distance between u and v, or 0 if there is no path between them. Compute the sum of values d(u, v) over all 1 ≤ u < v ≤ n.
The gcd (greatest common divisor) of two positive integers is the maximum positive integer that divides both of the integers.
Input
Single integer n (1 ≤ n ≤ 107).
Output
Print the sum of d(u, v) over all 1 ≤ u < v ≤ n.
Examples
Input
6
Output
8
Input
10
Output
44
Note
All shortest paths in the first example:
* <image>
* <image>
* <image>
* <image>
* <image>
* <image>
There are no paths between other pairs of vertices.
The total distance is 2 + 1 + 1 + 2 + 1 + 1 = 8.
Submitted Solution:
```
import math
n = int(input())
gr = []
kek = 0
for i in range(n):
gr.append([])
for ii in range(n):
if i == ii: gr[i].append(0)
elif math.gcd(i+1,ii+1) != 1: gr[i].append(1)
else: gr[i].append(0)
for i in range(1,n-1):
for k in range(i+1,n):
if gr[i][k] == 0:
for j in range(1,n):
if gr[i][j] and gr[j][k]:
kek += 2
break
else:
for j in range(1,n):
for t in range(1,n):
if gr[i][j] and gr[j][t] and gr[t][k]:
kek += 3
break
else:
kek += 1
print(kek)
``` | instruction | 0 | 56,127 | 13 | 112,254 |
No | output | 1 | 56,127 | 13 | 112,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer n. Let's build a graph on vertices 1, 2, ..., n in such a way that there is an edge between vertices u and v if and only if <image>. Let d(u, v) be the shortest distance between u and v, or 0 if there is no path between them. Compute the sum of values d(u, v) over all 1 ≤ u < v ≤ n.
The gcd (greatest common divisor) of two positive integers is the maximum positive integer that divides both of the integers.
Input
Single integer n (1 ≤ n ≤ 107).
Output
Print the sum of d(u, v) over all 1 ≤ u < v ≤ n.
Examples
Input
6
Output
8
Input
10
Output
44
Note
All shortest paths in the first example:
* <image>
* <image>
* <image>
* <image>
* <image>
* <image>
There are no paths between other pairs of vertices.
The total distance is 2 + 1 + 1 + 2 + 1 + 1 = 8.
Submitted Solution:
```
n=int(input())
s = 0
b = [1] * n
c = [0] * n
c[0] = 1
nu = 0
for i in range(1,n):
if c[i] == 0:
j = i
while j < n:
if c[j] == 0:
c[j] = i+1
j += i+1
j = i
st = 1
stt = 1
sst = (i+1) ** 2
while j < n:
b[j] *= i
if st % (i+1) == 0:
stt *= i+1
if j+1 == sst:
b[j] *= stt
sst = sst *(i+1)
k = j+sst
while k < n:
b[k] *= stt
k += sst
j += i+1
st += 1
sdt=0
ed = (n*(n-1)) // 2
for i in range(1,n):
sdt += b[i]
print(ed)
ed -= sdt
sdv = 0
for i in range(2,n+1):
for j in range(i+1,n+1):
if c[i-1]*c[j-1] <= n:
sdv += 1
## print(str(i)+' '+str(j))
else:
if max(c[i-1],c[j-1])*2 > n:
nu += 1
sdv = sdv - ed
st = sdt - (n-1) - nu - sdv
s = ed + sdv*2 + st*3
print(str(ed)+' '+str(sdv)+' '+str(st)+' '+str(nu)+' '+str(sdt))
print(' '.join(map(str,c)))
print(' '.join(map(str,b)))
print(s)
``` | instruction | 0 | 56,128 | 13 | 112,256 |
No | output | 1 | 56,128 | 13 | 112,257 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,243 | 13 | 112,486 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
MOD=10**9+7
def dot(a, b):
res = [[0] * n for _ in range(n)]
for i in range(n):
for k, aik in enumerate(a[i]):
for j,bkj in enumerate(b[k]):
res[i][j] += aik*bkj
res[i][j] %= MOD
return res
n,k = [int(i) for i in input().split()]
A = [[int(i) for i in input().split()] for _ in range(n)]
ans = [[0] * n for _ in range(n)]
for i in range(n):
ans[i][i] = 1
while k:
k, i = divmod(k, 2)
if i:
ans = dot(ans, A)
A = dot(A, A)
res = 0
for i in range(n):
for j in range(n):
res += ans[i][j]
res %= MOD
print(res)
``` | output | 1 | 56,243 | 13 | 112,487 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,244 | 13 | 112,488 |
"Correct Solution:
```
mod = 10**9+7
def mat_mul(A, B):
# A,Bは正方行列とする
C = [[0]*N for _ in range(N)]
for i in range(N):
for j in range(N):
C[i][j] = sum(A[i][k] * B[k][j] for k in range(N)) % mod
return C
def matpow(A, n):
B = [[0]*N for _ in range(N)]
for i in range(N):
B[i][i] = 1
while n > 0:
if n & 1:
B = mat_mul(A, B)
A = mat_mul(A, A)
n //= 2
return B
if __name__ == "__main__":
N, K = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(N)]
A = matpow(A, K)
ans = 0
for a in A:
ans = (ans + sum(a)) % mod
print(ans)
``` | output | 1 | 56,244 | 13 | 112,489 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,245 | 13 | 112,490 |
"Correct Solution:
```
import sys
stdin = sys.stdin
def ni(): return int(ns())
def na(): return list(map(int, stdin.readline().split()))
def naa(N): return [na() for _ in range(N)]
def ns(): return stdin.readline().rstrip() # ignore trailing spaces
def matrix_cal(A, B, mod):
a = len(A)
b = len(A[0])
c = len(B)
d = len(B[0])
assert(b == c)
ans = [[0] * d for _ in range(a)]
for aa in range(a):
for dd in range(d):
v = 0
for bb in range(b):
v = (v + A[aa][bb] * B[bb][dd]) % mod
ans[aa][dd] = v
return ans
def matrix_pow(A, n, mod=0):
if n == 1:
return A
a = matrix_pow(A, n//2, mod)
ans = matrix_cal(a, a, mod)
if n % 2 == 1:
ans = matrix_cal(ans, A, mod)
return ans
N, K = na()
A = naa(N)
mod = 10 ** 9 + 7
B = matrix_pow(A, K, mod)
ans = 0
for i in range(N):
for j in range(N):
ans = (ans + B[i][j]) % mod
print(ans)
``` | output | 1 | 56,245 | 13 | 112,491 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,246 | 13 | 112,492 |
"Correct Solution:
```
n, kk = map(int, input().split())
AA = [list(map(int, input().split())) for _ in range(n)]
mod = 10 ** 9 + 7
# nxnの行列積
def matrix_product(A, B):
res = [[0] * n for _ in range(n)]
for i in range(n):
for k in range(n):
for j in range(n):
res[i][j] += A[i][k] * B[k][j]
res[i][j] %= mod
return res
def matrix_exponents(A, t):
res = [[0] * n for _ in range(n)]
for i in range(n):
res[i][i] = 1
while t:
if t & 1 == 1:
res = matrix_product(res, A)
A = matrix_product(A, A)
t //= 2
return res
temp = matrix_exponents(AA, kk)
ans = 0
for i in range(n):
for j in range(n):
ans += temp[i][j]
ans %= mod
print(ans)
``` | output | 1 | 56,246 | 13 | 112,493 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,247 | 13 | 112,494 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007
n, k = LI()
G = LIR(n)
def mat_mul(a, b):
I, J, K = len(a), len(b[0]), len(b)
c = [[0] * J for _ in range(I)]
for i in range(I):
for j in range(J):
for k in range(K):
c[i][j] += a[i][k] * b[k][j]
c[i][j] %= mod
return c
def mat_pow(x, n):
y = [[0] * len(x) for _ in range(len(x))]
for i in range(len(x)):
y[i][i] = 1
while n > 0:
if n & 1:
y = mat_mul(x, y)
x = mat_mul(x, x)
n >>= 1
return y
ans = 0
GG = mat_pow(G, k)
for i in range(n):
for j in range(n):
ans += GG[i][j]
print(ans % mod)
``` | output | 1 | 56,247 | 13 | 112,495 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,248 | 13 | 112,496 |
"Correct Solution:
```
p = 10**9+7
def pow(X,k):
if k==0:
return [[1 if i==j else 0 for j in range(N)] for i in range(N)]
if k==1:
return X
if k%2==0:
Y = pow(X,k//2)
Z = [[0 for _ in range(N)] for _ in range(N)]
for i in range(N):
for j in range(N):
for m in range(N):
Z[i][j] = (Z[i][j]+Y[i][m]*Y[m][j])%p
return Z
else:
Y = pow(X,(k-1)//2)
Z = [[0 for _ in range(N)] for _ in range(N)]
for i in range(N):
for j in range(N):
for m in range(N):
Z[i][j] = (Z[i][j]+Y[i][m]*Y[m][j])%p
Z1 = [[0 for _ in range(N)] for _ in range(N)]
for i in range(N):
for j in range(N):
for m in range(N):
Z1[i][j] = (Z1[i][j]+Z[i][m]*X[m][j])%p
return Z1
N,K = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
X = pow(A,K)
cnt = 0
for i in range(N):
for j in range(N):
cnt = (cnt+X[i][j])%p
print(int(cnt))
``` | output | 1 | 56,248 | 13 | 112,497 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,249 | 13 | 112,498 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def matDot(A,B,MOD):
N,M,L = len(A),len(A[0]),len(B[0])
res = [[0]*L for i in range(N)]
for i in range(N):
for j in range(L):
s = 0
for k in range(M):
s = (s + A[i][k]*B[k][j]) % MOD
res[i][j] = s
return res
def matPow(A,x,MOD):
N = len(A)
res = [[0]*N for i in range(N)]
for i in range(N):
res[i][i] = 1
for i in range(x.bit_length()):
if (x>>i) & 1:
res = matDot(res,A,MOD)
A = matDot(A,A,MOD)
return res
n,k = map(int,input().split())
mod = 10**9+7
a = []
for i in range(n):
a.append(tuple(map(int,input().split())))
b = matPow(a,k,mod)
ans = 0
for i in range(n):
for j in range(n):
ans = (ans+b[i][j])%mod
print(ans)
``` | output | 1 | 56,249 | 13 | 112,499 |
Provide a correct Python 3 solution for this coding contest problem.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352 | instruction | 0 | 56,250 | 13 | 112,500 |
"Correct Solution:
```
n,k = map(int,input().split())
g = [list(map(int,input().split())) for i in range(n)]
mod = 10**9+7
ans = 0
def mul(x,y):
res = []
for i in range(n):
tmp = []
for j in range(n):
ko = 0
for k in range(n):
ko += x[i][k] * y[k][j]
ko %=mod
tmp.append(ko)
res.append(tmp)
return res
while k>0:
if k &1 ==1:
if type(ans) is int:
ans = g
else:
ans = mul(g,ans)
g = mul(g,g)
k >>= 1
res = 0
for i in range(n):
for j in range(n):
res += ans[i][j]
res %= mod
print(res)
``` | output | 1 | 56,250 | 13 | 112,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
N,K = map(int,input().split())
G = [list(map(int,input().split())) for _ in range(N)]
#print(G)
MOD = 10**9+7
def dot(a,b):
res = [[0] * N for _ in range(N)]
for i in range(N):
for j in range(N):
res[i][j] = sum(a[i][k] * b[k][j] for k in range(N)) % MOD
return res
ans = [[0] * N for _ in range(N)]
for i in range(N):
ans[i][i] = 1
while(K>0):
K,x = divmod(K,2)
if x == 1:
ans = dot(ans,G)
G = dot(G,G)
print(sum(sum(a) % MOD for a in ans) % MOD)
``` | instruction | 0 | 56,251 | 13 | 112,502 |
Yes | output | 1 | 56,251 | 13 | 112,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
mod = 10**9 + 7
N, K = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
def dot(X, Y):
x = len(X)
y = len(Y[0])
z = len(Y)
C = [[0] * y for _ in range(x)]
for i in range(x):
for j in range(y):
for k in range(z):
C[i][j] += X[i][k] * Y[k][j]
C[i][j] %= mod
return C
B = [[0] * N for _ in range(N)]
for k in range(N):
B[k][k] = 1
while K > 0:
if K & 1:
B = dot(B, A)
A = dot(A, A)
K >>= 1
ans = 0
for i in range(N):
for j in range(N):
ans += B[i][j]
ans %= mod
print(ans)
``` | instruction | 0 | 56,252 | 13 | 112,504 |
Yes | output | 1 | 56,252 | 13 | 112,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
#!/mnt/c/Users/moiki/bash/env/bin/python
N,K = map(int, input().split())
A = [ [int(j) for j in input().split()] for i in range(N)]
# I = [ [j for j in range(1,N+1) if Ai[j]] for Ai in A]
MOD = 10**9 + 7
# dp = [0] * (N+1)
dp = [ [0] * (N+1) for _ in range(N+1)]
S = [[1]*N]
KDIG = [0]*70
for i in range(1,61):
if K % (1<<i) :
KDIG[i-1] = 1
K -= K%(1<<i)
def prod(A,B,k,l,m):#A:k*l,B:l*m
C=[[None for i in range(m)] for j in range(k)]
for i in range(k):
for j in range(m):
ANS=0
for pl in range(l):
ANS=(ANS+A[i][pl]*B[pl][j])%MOD
C[i][j]=ANS
return C
for i in range(60):
if KDIG[i] == 1:
S = prod(S,A,1,N,N)
A = prod(A,A,N,N,N)
print(sum(S[0])%MOD)
# for i in range(1,N+1):
# for j in range(1,N+1):
# # dp[i][j] = np.linalg.matrix_power(A, K)[i-1][j-1].item()
# # ans += dp[i][j] % MOD
# # ans += np.linalg.matrix_power(A, K)[i-1][j-1].item() % MOD
# # ans += M[i-1,j-1].item() % MOD
# ans += M[i-1][j-1] % MOD
# print(ans%MOD)
``` | instruction | 0 | 56,253 | 13 | 112,506 |
Yes | output | 1 | 56,253 | 13 | 112,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
mod = 10**9+7
n, K = map(int, input().split())
a = [[int(x) for x in input().split()] for _ in range(n)]
A = [a]
i = 1
while 2**i <= K:
mtrix = []
for j in range(n):
raw = []
for l in range(n):
tmp = 0
for k in range(n):
tmp += A[-1][j][k] * A[-1][k][l]
raw.append(tmp % mod)
mtrix.append(raw)
A.append(mtrix)
i += 1
bit_max = i
ans = [[[0]*n for _ in range(n)]]
for i in range(n):
for j in range(n):
if i == j:
ans[-1][i][j] = 1
for i in range(bit_max):
if (K >> i) & 1:
mtrix = []
for j in range(n):
raw = []
for l in range(n):
tmp = 0
for k in range(n):
tmp += ans[-1][j][k] * A[i][k][l]
raw.append(tmp % mod)
mtrix.append(raw)
ans.append(mtrix)
ans = sum(map(lambda x: sum(x) % mod, ans[-1]))
print(ans % mod)
``` | instruction | 0 | 56,254 | 13 | 112,508 |
Yes | output | 1 | 56,254 | 13 | 112,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
n,k = map(int,input().split())
g = [list(map(int,input().split())) for i in range(n)]
mod = 10**9+7
import numpy as np
gyou = np.matrix(g)
ans = np.eye(n).astype(int)
ans = np.matrix(ans)
while k:
if k &1 ==1:
ans = np.dot(ans,gyou)
gyou = np.dot(gyou,gyou)
k >>= 1
ans %=mod
gyou %= mod
print(np.sum(ans)%mod)
``` | instruction | 0 | 56,255 | 13 | 112,510 |
No | output | 1 | 56,255 | 13 | 112,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
import numpy as np
import sys
input = sys.stdin.readline
mod = 10**9+7
N, K = map(int, input().split())
a = np.array([list(map(int, input().split())) for _ in range(N)], dtype=np.int64)
ans = np.identity(N, dtype=np.int64)
while K > 0:
if K & 1 == 1:
ans = np.matmul(ans, a) % mod
a = np.matmul(a, a) % mod
K >>= 1
print(np.sum(ans) % mod)
``` | instruction | 0 | 56,256 | 13 | 112,512 |
No | output | 1 | 56,256 | 13 | 112,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
from collections import deque
from collections import Counter
from itertools import product, permutations,combinations
from operator import itemgetter
from heapq import heappop, heappush
from bisect import bisect_left, bisect_right, bisect
#pypyではscipy, numpyは使えない
#from scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra, bellman_ford, johnson, minimum_spanning_tree
#from scipy.sparse import csr_matrix, coo_matrix, lil_matrix
import numpy as np
from fractions import gcd
from math import ceil,floor, sqrt, cos, sin, pi, factorial
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
#文字列のときはうまく行かないのでコメントアウトすること
sys.setrecursionlimit(10**9)
INF = float('inf')
MOD = 10**9 +7
def main():
N, K = map(int, readline().split())
A = np.array(read().split(), np.int64).reshape(N,N)
def pow_array(a,k):#行列aのk乗の計算
ans = np.eye(N)
while k:
if k%2:
ans = np.dot(ans%MOD, a%MOD)
a = np.dot(a%MOD, a%MOD)
k>>=1
return ans%MOD
anser = pow_array(A,K).astype('int64').sum()
print(anser)
if __name__ == '__main__':
main()
``` | instruction | 0 | 56,257 | 13 | 112,514 |
No | output | 1 | 56,257 | 13 | 112,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.
For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j. If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.
Find the number of different directed paths of length K in G, modulo 10^9 + 7. We will also count a path that traverses the same edge multiple times.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 10^{18}
* a_{i, j} is 0 or 1.
* a_{i, i} = 0
Input
Input is given from Standard Input in the following format:
N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}
Output
Print the number of different directed paths of length K in G, modulo 10^9 + 7.
Examples
Input
4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0
Output
6
Input
3 3
0 1 0
1 0 1
0 0 0
Output
3
Input
6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0
Output
1
Input
1 1
0
Output
0
Input
10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0
Output
957538352
Submitted Solution:
```
import numpy as np
def main():
MOD = int(1e9 + 7)
n, k = map(int, input().split())
can = [[int(c) for c in input().split()] for _ in range(n)]
can = np.array(can)
answer = np.eye(n)
while k:
if k % 2:
answer = np.dot(answer, can) % MOD
can = np.dot(can, can) % MOD
k //= 2
total = 0
for i in range(n):
for j in range(n):
total += answer[i,j]
if total >= MOD:
total -= MOD
print(total)
if __name__ == "__main__":
main()
``` | instruction | 0 | 56,258 | 13 | 112,516 |
No | output | 1 | 56,258 | 13 | 112,517 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,379 | 13 | 112,758 |
"Correct Solution:
```
# coding=utf-8
class Tree():
def __init__(self):
self.root = None
def insert(self, key):
node = Node(key)
flag = False
if self.root == None:
self.root = node
else:
search_node = self.root
while search_node:
final_node = search_node
if final_node.key < key:
search_node = final_node.right
flag = 'right'
else:
search_node = final_node.left
flag = 'left'
node.parent = final_node
if flag == 'right':
final_node.right = node
elif flag == 'left':
final_node.left = node
def print_tree(self):
prewalked = self.root.prewalk()
inwalked = self.root.inwalk()
print(' ', end = '')
print(*inwalked)
print(' ', end = '')
print(*prewalked)
class Node():
def __init__(self, key):
self.key = key
self.parent = None
self.left = None
self.right = None
def prewalk(self):
result = []
result += [self.key] # ??¢?´¢????????? #
if self.left:
result += self.left.prewalk()
if self.right:
result += self.right.prewalk()
#print("self.key:", self.key, result) #debugger
return result
def inwalk(self):
result = []
if self.left:
result += self.left.inwalk()
result += [self.key] # ??¢?´¢????????? #
if self.right:
result += self.right.inwalk()
#print("self.key:", self.key, result) #debugger
return result
n = int(input())
tree = Tree()
for _ in range(n):
order, *key = input().split()
if order == 'insert':
tree.insert(int(key[0]))
else:
tree.print_tree()
``` | output | 1 | 56,379 | 13 | 112,759 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,380 | 13 | 112,760 |
"Correct Solution:
```
import sys
class Node:
__slots__ = ['key', 'left', 'right']
def __init__(self, key):
self.key = key
self.left = self.right = None
root = None
def insert(key):
global root
x, y = root, None
while x is not None: x, y = x.left if key < x.key else x.right, x
if y is None: root = Node(key)
elif key < y.key: y.left = Node(key)
else: y.right = Node(key)
def inorder(node):
return inorder(node.left) + f' {node.key}' + inorder(node.right) if node else ''
def preorder(node):
return f' {node.key}' + preorder(node.left) + preorder(node.right) if node else ''
input()
for e in sys.stdin:
if e[0] == 'i': insert(int(e[7:]))
else: print(inorder(root)); print(preorder(root))
``` | output | 1 | 56,380 | 13 | 112,761 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,381 | 13 | 112,762 |
"Correct Solution:
```
class Node:
def __init__(self, key, parent=None, left=None, right=None):
self.key = key
self.parent = parent
self.left = left
self.right = right
class Tree:
root = None
preorder_list = []
inorder_list = []
def __init__(self):
pass
def insert(self, node):
y = None
x = self.root
while x is not None:
y = x
if node.key < x.key:
x = x.left
else:
x = x.right
node.parent = y
if y == None:
self.root = node
elif node.key < y.key:
y.left = node
else:
y.right = node
def display(self):
#print(self.root, self.root.key, self.root.parent, self.root.left, self.root.right)
self.inorder(self.root)
self.preorder(self.root)
print(' '+' '.join(map(str, self.inorder_list)))
print(' '+' '.join(map(str, self.preorder_list)))
pass
def preorder(self, node):
if node == None:
return
self.preorder_list.append(node.key)
if node.left is not None:
self.preorder(node.left)
if node.right is not None:
self.preorder(node.right)
def inorder(self, node):
if node == None:
return
if node.left is not None:
self.inorder(node.left)
self.inorder_list.append(node.key)
if node.right is not None:
self.inorder(node.right)
n = int(input())
command_list = [list(input().split()) for _ in range(n)]
tree = Tree()
for command in command_list:
if command[0] == 'insert':
node = Node(int(command[1]))
tree.insert(node)
#print(node, node.key, node.parent, node.left, node.right)
if command[0] == 'print':
tree.display()
tree.preorder_list = []
tree.inorder_list = []
``` | output | 1 | 56,381 | 13 | 112,763 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,382 | 13 | 112,764 |
"Correct Solution:
```
n = int(input())
class Node:
def __init__(self):
self.key = -1
self.parent_id = -1
self.left_id = -1
self.right_id = -1
def insert(nodes,z):
y = -1
root_id = -1
for i in range(len(nodes)):
if nodes[i].parent_id == -1:
root_id = i
break
x = root_id
while x != -1:
y = x
if z.key < nodes[x].key:
x = nodes[x].left_id
else:
x = nodes[x].right_id
z.parent_id = y
nodes.append(z)
z_id = len(nodes) - 1
if y == -1: pass
elif z.key < nodes[y].key:
nodes[y].left_id = z_id
else:
nodes[y].right_id = z_id
def pre(node_id):
pre_list.append(nodes[node_id].key)
left = nodes[node_id].left_id
right = nodes[node_id].right_id
if left != -1:
pre(left)
if right != -1:
pre(right)
def Ino(node_id):
if nodes[node_id].left_id != -1 :
Ino(nodes[node_id].left_id)
ino_list.append(nodes[node_id].key)
if nodes[node_id].right_id != -1:
Ino(nodes[node_id].right_id)
nodes = []
for i in range(n):
cmd = input()
op = cmd[0]
if op == "i":
z = Node()
number = int(cmd[7:])
z.key = number
insert(nodes,z)
if op =="p":
pre_list = []
ino_list = []
for i in range(len(nodes)):
if nodes[i].parent_id == -1:
Ino(i)
pre(i)
break
pre_list = list(map(str, pre_list))
ino_list = list(map(str, ino_list))
print(" ", end="")
print(' '.join(ino_list))
print(" ", end="")
print(' '.join(pre_list))
``` | output | 1 | 56,382 | 13 | 112,765 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,383 | 13 | 112,766 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
print = sys.stdout.write
class Node:
__slots__ = ["data", "left", "right"]
def __init__(self, data):
self.data = data
self.left = None
self.right = None
class BinarySearchTree:
__slots__ = ["root"]
def __init__(self):
self.root = None
def insert(self, data):
parent = self.root
if parent is None:
self.root = Node(data)
return
while parent:
parent_old = parent
parent = parent.left if data < parent.data else parent.right
if data < parent_old.data:
parent_old.left = Node(data)
else:
parent_old.right = Node(data)
def print_preorder(node):
print(" {}".format(node.data))
if node.left:
print_preorder(node.left)
if node.right:
print_preorder(node.right)
def print_inorder(node):
if node.left:
print_inorder(node.left)
print(" {}".format(node.data))
if node.right:
print_inorder(node.right)
if __name__ == "__main__":
n = int(input())
binary_search_tree = BinarySearchTree()
for _ in range(n):
operation, *num = input().split()
if num:
binary_search_tree.insert(int(num[0]))
elif binary_search_tree.root is not None:
print_inorder(binary_search_tree.root)
print("\n")
print_preorder(binary_search_tree.root)
print("\n")
``` | output | 1 | 56,383 | 13 | 112,767 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,384 | 13 | 112,768 |
"Correct Solution:
```
class BinarySearchTree:
root = None
def insert(self, k):
y = None
x = self.root
z = Node(k)
while x:
y = x
if z.k < x.k:
x = x.l
else:
x = x.r
z.p = y
if y == None:
self.root = z
else:
if z.k < y.k:
y.l = z
else:
y.r = z
def preParse(self, u):
if u == None:
return
print("", str(u.k), end = "")
self.preParse(u.l)
self.preParse(u.r)
def inParse(self, u):
if u == None:
return
self.inParse(u.l)
print("", str(u.k), end = "")
self.inParse(u.r)
class Node:
def __init__(self, k):
self.k = k
self.p = None
self.l = None
self.r = None
if __name__ == '__main__':
n = int(input().rstrip())
x = BinarySearchTree()
for i in range(n):
com = input().rstrip()
if com.startswith("insert"):
k = int(com.split(" ")[1])
x.insert(k)
else:
x.inParse(x.root)
print()
x.preParse(x.root)
print()
``` | output | 1 | 56,384 | 13 | 112,769 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,385 | 13 | 112,770 |
"Correct Solution:
```
"""
二分探索木を構成する
"""
import copy
class Node:
def __init__(self, node_id, parent=None, left=None, right=None):
self.node_id = node_id
self.parent = parent
self.left = left
self.right = right
class Tree:
def insert_root_node(self, new_id):
global root_node #途中でインスタンス化すると水の泡なんですよ
root_node = Node(new_id)
def insert_node(self, root_id, new_id): #new_idは追加するノード
compare_node = root_node #はじめはrootから比較スタート
new_node = Node(new_id) #インスタンス化
while compare_node != None: #compare_nodeが存在するとき
#この下の一行は最重要アイディア
#親の可能性がある時，とりあえず代入して情報を記録しておく
new_node.parent = compare_node
#右か左か
if compare_node.node_id < new_id:
compare_node = compare_node.right
else:
compare_node = compare_node.left
#compare_node == Noneの時，上の２つの分岐のどっちからきたかわからない
#そのため，親と比較し直す必要がある
if new_node.parent.node_id < new_id:
new_node.parent.right = Node(new_id)
else:
new_node.parent.left = Node(new_id)
def inParse(tmp_node):
if not tmp_node:
return
inParse(tmp_node.left)
print(' '+str(tmp_node.node_id), end='')
inParse(tmp_node.right)
def preParse(tmp_node):
if not tmp_node:
return
print(' '+str(tmp_node.node_id), end='')
preParse(tmp_node.left)
preParse(tmp_node.right)
tree = Tree()
n = int(input())
root_id = 0
root_node = None
for i in range(n):
order = input().split()
if root_node == None: #何も入ってない時
root_id = int(order[1])
tree.insert_root_node(int(order[1]))
elif order[0][0] == 'i':
new_id = int(order[1])
tree.insert_node(root_id, new_id)
elif order[0][0] == 'p':
node = copy.copy(root_node)
inParse(node)
print()
node = copy.copy(root_node)
preParse(node)
print()
``` | output | 1 | 56,385 | 13 | 112,771 |
Provide a correct Python 3 solution for this coding contest problem.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88 | instruction | 0 | 56,386 | 13 | 112,772 |
"Correct Solution:
```
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_8_A
#????????????
def insert(root, insert_node):
focus_node = root
parent = None
while not focus_node == None:
parent = focus_node
if focus_node["data"] > insert_node["data"]:
focus_node = focus_node["left"]
else:
focus_node = focus_node["right"]
if parent["data"] > insert_node["data"]:
parent["left"] = insert_node
else:
parent["right"] = insert_node
def get_preorder(node):
if node == None:
return []
r = []
r.append(str(node["data"]))
r.extend(get_preorder(node["left"]))
r.extend(get_preorder(node["right"]))
return r
def get_inorder(node):
if node == None:
return []
r = []
r.extend(get_inorder(node["left"]))
r.append(str(node["data"]))
r.extend(get_inorder(node["right"]))
return r
def print_tree(root):
print(" " + " ".join(get_inorder(root)))
print(" " + " ".join(get_preorder(root)))
def main():
n_line = int(input())
input_list = [input() for i in range(n_line)]
root = {"left":None, "right": None, "data":int(input_list[0].split()[1])}
for line in input_list[1:]:
if line == "print":
print_tree(root)
else:
node = {"left":None, "right": None, "data":int(line.split()[1])}
insert(root, node)
if __name__ == "__main__":
main()
``` | output | 1 | 56,386 | 13 | 112,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
import sys
class Node:
__slots__ = ['key', 'left', 'right']
def __init__(self, key):
self.key = key
self.left = self.right = None
root = None
def insert(key):
global root
x, y = root, None
while x:
y = x
if key < x.key:
x = x.left
else:
x = x.right
if y is None:
root = Node(key)
elif key < y.key:
y.left = Node(key)
else:
y.right = Node(key)
def inorder(node):
return inorder(node.left) + f' {node.key}' + inorder(node.right) if node else ''
def preorder(node):
return f' {node.key}' + preorder(node.left) + preorder(node.right) if node else ''
n = int(input())
for e in sys.stdin:
if e[0] == 'i':
insert(int(e[7:]))
else:
print(inorder(root))
print(preorder(root))
``` | instruction | 0 | 56,387 | 13 | 112,774 |
Yes | output | 1 | 56,387 | 13 | 112,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
import sys
class Node():
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.parent = None
def preorder(self):
nodeList = [self.key]
if self.left:
nodeList += self.left.preorder()
if self.right:
nodeList += self.right.preorder()
return nodeList
def inorder(self):
nodeList = []
if self.left:
nodeList += self.left.inorder()
nodeList += [self.key]
if self.right:
nodeList += self.right.inorder()
return nodeList
class BinaryTree():
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y is None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def print(self):
print(" " + " ".join(map(str, self.root.inorder())))
print(" " + " ".join(map(str, self.root.preorder())))
if __name__ == "__main__":
tree = BinaryTree()
n = int(sys.stdin.readline())
for inp in sys.stdin.readlines():
inp = inp.split()
if inp[0] == "insert":
tree.insert(int(inp[1]))
else:
tree.print()
``` | instruction | 0 | 56,388 | 13 | 112,776 |
Yes | output | 1 | 56,388 | 13 | 112,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
import sys
class Node:
def __init__(self, value, p = None, l = None, r = None):
self.key = value
self.p = p
self.left = l
self.right = r
def getroot(x):
if x.p != None:
return getroot(x.p)
return x
def preorder(x, A):
if x == None:
return
A.append(x.key)
preorder(x.left, A)
preorder(x.right, A)
def inorder(x, A):
if x == None:
return
inorder(x.left, A)
A.append(x.key)
inorder(x.right, A)
def postorder(x, A):
if x == None:
return
postorder(x.left, A)
postorder(x.right, A)
A.append(x.key)
def ptree(root):
pre = [""]
ino = [""]
preorder(root, pre)
inorder(root, ino)
print(" ".join(map(str,ino)))
print(" ".join(map(str,pre)))
def insert(Tree, root, z):
y = None # x ??????
x = root # 'T ??????'
while x != None:
y = x # ???????¨????
if z.key < x.key:
x = x.left # ?????????????§????
else:
x = x.right # ?????????????§????
z.p = y
Tree.append(z)
if y == None: # T ???????????´???
return z
elif z.key < y.key:
y.left = z # z ??? y ?????????????????????
else:
y.right = z # z ??? y ?????????????????????
return root
def main():
""" ????????? """
num = int(input().strip())
istr = sys.stdin.read()
cmds = list(istr.splitlines())
Tree = []
root = None
for i in range(num):
cmd = cmds[i][0]
if cmd == "i":
n = Node(int(cmds[i][7:]))
root = insert(Tree, root, n)
elif cmd == "p":
ptree(root)
if __name__ == '__main__':
main()
``` | instruction | 0 | 56,389 | 13 | 112,778 |
Yes | output | 1 | 56,389 | 13 | 112,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
class Node:
def __init__(self,key,left,right,parent):
self.key = key
self.left = left
self.right = right
self.p = parent
def __repr__(self):
return str(self.key)
def insert(T, z):
y = None
x = T
while x != None:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.p = y
#print(x,y,z)
if y == None:
#print("Y==None")
T = z
print(T)
elif z.key<y.key:
#print("z.key<y.key")
y.left = z
else:
#print("z.key<y.key")
y.right = z
def printPreorder(x):
if x == None:
return
printPreorder(x.left)
print(end=' ')
print(x,end='')
printPreorder(x.right)
def printMidorder(x):
if x == None:
return
print(end=' ')
print(x,end='')
printMidorder(x.left)
printMidorder(x.right)
def printal(x):
printPreorder(x)
print()
printMidorder(x)
print()
N = int(input())
Q = input().split()
T = Node(int(Q[1]),None,None,None)
for i in range(N-1):
Q = input().split()
if Q[0] == "print":
printal(T)
continue
insert(T,Node(int(Q[1]),None,None,None))
``` | instruction | 0 | 56,390 | 13 | 112,780 |
Yes | output | 1 | 56,390 | 13 | 112,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
#!/usr/bin/env python3
import sys
#run by python3
class Node:
def __init__(self, key):
self.key = key
self.right = None
self.left = None
def insert(T, z): #z is node which you would like to insert
#initialization
y = None #parent of x
x = T
#search
while(x!=None):
y = x
if(z.key < x.key):
x = x.left
else:
x = x.right
#insert
if(y==None):
T = z
elif(z.key < y.key):
y.left = z
else:
y.right = z
return T
def inorder_walk(node):
if node.left:
inorder_walk(node.left)
print ('', node.key, end='')
if node.right:
inorder_walk(node.right)
def preorder_walk(node):
print ('', node.key,end='')
if node.left:
preorder_walk(node.left)
if node.right:
preorder_walk(node.right)
if __name__ == '__main__':
T = None
num = int(sys.stdin.readline())
for _ in range(num):
cmd = sys.stdin.readline()
if(cmd.startswith('i')):
n = Node(int(cmd[7:]))
T = insert(T, n)
else:
inorder_walk(T)
print()
preorder_walk(T)
print()
``` | instruction | 0 | 56,391 | 13 | 112,782 |
No | output | 1 | 56,391 | 13 | 112,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
class NullNode():
def __init__(self):
self.id = None
class Node():
def __init__(self, id):
self.id = id
self.parent = NullNode()
self.left = NullNode()
self.right = NullNode()
def preorder(node, out_list=[]):
if node.id is not None:
out_list.append(str(node.id))
out_list = preorder(node.left, out_list)
out_list = preorder(node.right, out_list)
return out_list
def inorder(node, out_list=[]):
if node.id is not None:
out_list = inorder(node.left, out_list)
out_list.append(str(node.id))
out_list = inorder(node.right, out_list)
return out_list
def insert(node, root_node):
temp_node_parent = root_node.parent
temp_node = root_node
while temp_node.id is not None:
temp_node_parent = temp_node
if node.id < temp_node.id:
temp_node = temp_node.left
else:
temp_node = temp_node.right
node.parent = temp_node_parent
if node.id < temp_node_parent.id:
temp_node_parent.left = node
else:
temp_node_parent.right = node
def find(id,root_node):
node = root_node
while node.id is not None:
if id < node.id:
node = node.left
elif id > node.id:
node = node.right
else:
print('yes')
break
else:
print('no')
def delete(id, root_node):
# 作成中
type = ''
node = root_node
node_parent = NullNode()
while node.id != id:
node_parent = node
if id < node.id:
node = node.left
type = 'L'
else:
node = node.right
type = 'R'
if node.right is not None:
node.right.parent = node_parent
elif node.left is not None:
node.left.parent = node_parent
else:
node.parent = NullNode()
if type == 'L':
node_parent.left = NullNode()
else:
node_parent.right = NullNode()
import sys
n = int(input())
input_lines = sys.stdin.readlines()
root_node = NullNode()
for i in range(n):
input_line = input_lines[i].split()
command = input_line[0]
if command == 'insert':
id = int(input_line[1])
id_node = Node(id)
if root_node.id is None:
root_node = id_node
else:
insert(id_node, root_node)
elif command == 'find':
id = int(input_line[1])
find(id, root_node)
elif command == 'delete':
id = int(input_line[1])
delete(id, root_node)
else:
inorder_list = inorder(root_node, [])
preorder_list = preorder(root_node, [])
print(' ' + ' '.join(inorder_list))
print(' ' + ' '.join(preorder_list))
``` | instruction | 0 | 56,392 | 13 | 112,784 |
No | output | 1 | 56,392 | 13 | 112,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
class Node:
def __init__(self,x):
self.key = x
self.left = None
self.right = None
def insert(node, x):
if node is None:
return Node(x)
elif x == node.key:
return node
elif x < node.key:
node.left = insert(node.left, x)
else:
node.right = insert(node.right, x)
return node
def preorder(node):
if node == None:
return None
l.append(node.key)
preorder(node.left)
preorder(node.right)
return l
def inorder(node):
if node == None:
return None
inorder(node.left)
l.append(node.key)
inorder(node.right)
return l
class BinaryTree:
def __init__(self):
self.root = None
def insert(self, x):
self.root = insert(self.root, x)
def output(self):
print(self.root.key)
print(self.root.left.key)
print(self.root.right.key)
l = []
tree = BinaryTree()
n = int(input())
for i in range(n):
order = list(map(str, input().split()))
if order[0] == "insert":
tree.insert(int(order[1]))
if order[0] == "print":
inorder(tree.root)
for i in range(len(l)):
if i != len(l)-1:
print(l[i],end=" ")
else:
print(l[i])
l.clear()
preorder(tree.root)
for i in range(len(l)):
if i != len(l)-1:
print(l[i],end=" ")
else:
print(l[i])
exit()
else:
continue
``` | instruction | 0 | 56,393 | 13 | 112,786 |
No | output | 1 | 56,393 | 13 | 112,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Search trees are data structures that support dynamic set operations including insert, search, delete and so on. Thus a search tree can be used both as a dictionary and as a priority queue.
Binary search tree is one of fundamental search trees. The keys in a binary search tree are always stored in such a way as to satisfy the following binary search tree property:
* Let $x$ be a node in a binary search tree. If $y$ is a node in the left subtree of $x$, then $y.key \leq x.key$. If $y$ is a node in the right subtree of $x$, then $x.key \leq y.key$.
The following figure shows an example of the binary search tree.
<image>
For example, keys of nodes which belong to the left sub-tree of the node containing 80 are less than or equal to 80, and keys of nodes which belong to the right sub-tree are more than or equal to 80. The binary search tree property allows us to print out all the keys in the tree in sorted order by an inorder tree walk.
A binary search tree should be implemented in such a way that the binary search tree property continues to hold after modifications by insertions and deletions. A binary search tree can be represented by a linked data structure in which each node is an object. In addition to a key field and satellite data, each node contains fields left, right, and p that point to the nodes corresponding to its left child, its right child, and its parent, respectively.
To insert a new value $v$ into a binary search tree $T$, we can use the procedure insert as shown in the following pseudo code. The insert procedure is passed a node $z$ for which $z.key = v$, $z.left = NIL$, and $z.right = NIL$. The procedure modifies $T$ and some of the fields of $z$ in such a way that $z$ is inserted into an appropriate position in the tree.
1 insert(T, z)
2 y = NIL // parent of x
3 x = 'the root of T'
4 while x ≠ NIL
5 y = x // set the parent
6 if z.key < x.key
7 x = x.left // move to the left child
8 else
9 x = x.right // move to the right child
10 z.p = y
11
12 if y == NIL // T is empty
13 'the root of T' = z
14 else if z.key < y.key
15 y.left = z // z is the left child of y
16 else
17 y.right = z // z is the right child of y
Write a program which performs the following operations to a binary search tree $T$.
* insert $k$: Insert a node containing $k$ as key into $T$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.
You should use the above pseudo code to implement the insert operation. $T$ is empty at the initial state.
Constraints
* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.
Input
In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$ or print are given.
Output
For each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.
Example
Input
8
insert 30
insert 88
insert 12
insert 1
insert 20
insert 17
insert 25
print
Output
1 12 17 20 25 30 88
30 12 1 20 17 25 88
Submitted Solution:
```
class BinaryTree:
class Node:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.parent = None
def __init__(self):
self.root = None
def insert(self, key):
if self.root is None:
self.root = self.Node(key)
else:
root = self.root
temp = None
while True:
temp = root
if key > root.key:
root = root.right
if root is None:
temp.right = self.Node(key)
break
else:
root = root.left
if root is None:
temp.left = self.Node(key)
break
def print_inorder(self, node):
if node is not None:
self.print_inorder(node.left)
print(node.key, end = " ")
self.print_inorder(node.right)
def print_preorder(self, node):
if node is not None:
print(node.key, end = " ")
self.print_preorder(node.left)
self.print_preorder(node.right)
def test_insert(self, keys):
for k in keys:
self.insert(k)
length = int(input())
b = BinaryTree()
for _ in range(length):
comm = input()
if comm[0] == "i":
com, num = comm.split(" ")
b.insert(int(num))
else:
b.print_inorder(b.root)
print(" ")
b.print_preorder(b.root)
#
# b.test_insert([30,88,12,1,20,17,25])
# b.print_inorder(b.root)
# print(" ")
# b.print_preorder(b.root)
``` | instruction | 0 | 56,394 | 13 | 112,788 |
No | output | 1 | 56,394 | 13 | 112,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps:
1. choose two leaves;
2. add the length of the simple path between them to the answer;
3. remove one of the chosen leaves from the tree.
Initial answer (before applying operations) is 0. Obviously after n - 1 such operations the tree will consist of a single vertex.
Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!
Input
The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree.
Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.
Output
In the first line print one integer number — maximal possible answer.
In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation.
See the examples for better understanding.
Examples
Input
3
1 2
1 3
Output
3
2 3 3
2 1 1
Input
5
1 2
1 3
2 4
2 5
Output
9
3 5 5
4 3 3
4 1 1
4 2 2 | instruction | 0 | 56,898 | 13 | 113,796 |
Tags: constructive algorithms, dfs and similar, graphs, greedy, trees
Correct Solution:
```
import sys
from collections import deque as dq
n = int(input())
ind = 0
inp = [int(x)-1 for line in sys.stdin.readlines() for x in line.split()]
coupl = [[] for _ in range(n)]
for _ in range(n-1):
a,b = inp[ind],inp[ind+1]
ind+=2
coupl[a].append(b)
coupl[b].append(a)
Q = dq()
found = [False]*n
master = 0
found[master] = True
dia1 = 0
Q.append(master)
while Q:
node = Q.popleft()
dia1 = node
for nei in coupl[node]:
if not found[nei]:
found[nei] = True
Q.append(nei)
dia2 = 0
Q.append((dia1,0))
dist1 = [0]*n
found = [False]*n
found[dia1] = True
while Q:
node,d = Q.popleft()
dia2 = node
dist1[node]=d
for nei in coupl[node]:
if not found[nei]:
found[nei] = True
Q.append((nei,d+1))
Q = []
Q.append((dia2,0))
dist2 = [0]*n
found = [False]*n
found[dia2] = True
while Q:
node,d = Q.pop()
dist2[node]=d
for nei in coupl[node]:
if not found[nei]:
found[nei] = True
Q.append((nei,d+1))
neigs = [0]*n
leaves = []
for i in range(n):
if i != dia1 and i != dia2 and len(coupl[i])==1:
leaves.append(i)
neigs[i]=len(coupl[i])
points = 0
lista = []
while leaves:
node = leaves.pop()
if dist1[node]<dist2[node]:
lista.append((dia2,node,node))
points += dist2[node]
else:
lista.append((dia1,node,node))
points += dist1[node]
for nei in coupl[node]:
neigs[nei]-=1
if neigs[nei]==1:
leaves.append(nei)
leaves.append(dia2)
while leaves:
node = leaves.pop()
lista.append((dia1,node,node))
points += dist1[node]
for nei in coupl[node]:
neigs[nei]-=1
if neigs[nei]==1:
leaves.append(nei)
print(points)
for l in lista:
a,b,c = l
print(a+1,b+1,c+1)
``` | output | 1 | 56,898 | 13 | 113,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps:
1. choose two leaves;
2. add the length of the simple path between them to the answer;
3. remove one of the chosen leaves from the tree.
Initial answer (before applying operations) is 0. Obviously after n - 1 such operations the tree will consist of a single vertex.
Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!
Input
The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree.
Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.
Output
In the first line print one integer number — maximal possible answer.
In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation.
See the examples for better understanding.
Examples
Input
3
1 2
1 3
Output
3
2 3 3
2 1 1
Input
5
1 2
1 3
2 4
2 5
Output
9
3 5 5
4 3 3
4 1 1
4 2 2 | instruction | 0 | 56,899 | 13 | 113,798 |
Tags: constructive algorithms, dfs and similar, graphs, greedy, trees
Correct Solution:
```
import sys
def main():
n = int(input())
edges = list(map(int, sys.stdin.read().split()))
tree_edges = dict()
for i in range(n):
tree_edges[i + 1] = set()
for i in range(0, len(edges) - 1, 2):
tree_edges[edges[i]].add(edges[i + 1])
tree_edges[edges[i + 1]].add(edges[i])
init_distants = [-1] * (n + 1)
queue = [1]
init_distants[1] = 0
while queue:
next_queue = []
for process in queue:
for next_vertex in tree_edges[process]:
if init_distants[next_vertex] == -1:
init_distants[next_vertex] = init_distants[process] + 1
next_queue.append(next_vertex)
queue = next_queue
head = init_distants.index(max(init_distants))
distants_from_head = [-1] * (n + 1)
queue = [head]
distants_from_head[head] = 0
while queue:
next_queue = []
for process in queue:
for next_vertex in tree_edges[process]:
if distants_from_head[next_vertex] == -1:
distants_from_head[next_vertex] = distants_from_head[process] + 1
next_queue.append(next_vertex)
queue = next_queue
tail = distants_from_head.index(max(distants_from_head))
distants_from_tail = [-1] * (n + 1)
queue = [tail]
distants_from_tail[tail] = 0
while queue:
next_queue = []
for process in queue:
for next_vertex in tree_edges[process]:
if distants_from_tail[next_vertex] == -1:
distants_from_tail[next_vertex] = distants_from_tail[process] + 1
next_queue.append(next_vertex)
queue = next_queue
path_len_sum = 0
removal_history = list()
process_queue = []
for vertex, adj in tree_edges.items():
if len(adj) == 1:
process_queue.append(vertex)
while process_queue:
next_queue = []
for leaf in process_queue:
if leaf == head or leaf == tail:
continue
if distants_from_tail[leaf] > distants_from_head[leaf]:
path_len_sum += distants_from_tail[leaf]
new_leaves = []
for w in tree_edges[leaf]:
tree_edges[w].remove(leaf)
if len(tree_edges[w]) == 1:
new_leaves.append(w)
next_queue.extend(new_leaves)
removal_history.append("{0} {1} {0}".format(leaf, tail))
else:
path_len_sum += distants_from_head[leaf]
new_leaves = []
for w in tree_edges[leaf]:
tree_edges[w].remove(leaf)
if len(tree_edges[w]) == 1:
new_leaves.append(w)
next_queue.extend(new_leaves)
removal_history.append("{0} {1} {0}".format(leaf, head))
process_queue = next_queue
process_queue = [tail]
while process_queue:
leaf = process_queue[0]
if leaf == head:
continue
path_len_sum += distants_from_head[leaf]
new_leaves = []
for w in tree_edges[leaf]:
tree_edges[w].remove(leaf)
if len(tree_edges[w]) == 1:
new_leaves.append(w)
process_queue = new_leaves
removal_history.append("{0} {1} {0}".format(leaf, head))
print(str(path_len_sum))
sys.stdout.write("\n".join(removal_history))
sys.stdout.write("\n")
main()
``` | output | 1 | 56,899 | 13 | 113,799 |
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