message stringlengths 2 22.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,537 | 1 | 111,074 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10000000)
MOD = 10 ** 9 + 7
INF = 10 ** 15
class UnionFind():
def __init__(self,n):
self.n = n
self.parents = [-1]*n
def find(self,x): #根を見つける、繋ぎ直す
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def unite(self,x,y): #x,yの含むグループを併合する
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x,y = y,x
self.parents[x] += self.parents[y]
self.parents[y] = x
def same(self,x,y):#xとyが同じグループにいるか判定
return self.find(x) == self.find(y)
def solve(N,M):
edges = [tuple(map(int,input().split())) for _ in range(M)]
edges.sort(key = lambda x:x[2])
uf = UnionFind(N)
ans = 0
for a,b,c in edges:
if uf.same(a,b):
continue
else:
uf.unite(a,b)
ans += c
print(ans)
def main():
while True:
N,M = map(int,input().split())
if N == 0:
return
solve(N,M)
if __name__ == '__main__':
main()
``` | output | 1 | 55,537 | 1 | 111,075 |
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,538 | 1 | 111,076 |
"Correct Solution:
```
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * (n+1)
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n+1) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
def kr(edge,N):
res = 0
node = []
G = UnionFind(N)
for cost, p, q, in edge:
if not G.same(p, q):
G.union(p, q)
res += cost
return res
def solve():
n,m=map(int,input().split())
if n==0 and m==0:
exit()
edge=[]
for _ in range(m):
a,b,cost=map(int,input().split())
edge.append((cost,a,b))
edge.sort()
ans=kr(edge,n)
print(ans)
return solve()
if __name__=="__main__":
solve()
``` | output | 1 | 55,538 | 1 | 111,077 |
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,539 | 1 | 111,078 |
"Correct Solution:
```
# AOJ 0180 Demolition of Bridges
# Python3 2018.6.22
# UNION-FIND library
MAX = 105
id, size = [0]*MAX, [0]*MAX
def init(n):
for i in range(n): id[i], size[i] = i, 1
def root(i):
while i != id[i]:
id[i] = id[id[i]]
i = id[i]
return i
def connected(p, q): return root(p) == root(q)
def unite(p, q):
i, j = root(p), root(q)
if i == j: return
if size[i] < size[j]:
id[i] = j
size[j] += size[i]
else:
id[j] = i
size[i] += size[j]
# UNION-FIND library
# 最小全域木。V:総ノード数、E:枝情報(a,b,cost)
def kruskal(V, edge):
ee = sorted(edge, key=lambda x:(x[2]))
init(V)
ans = 0
for e in ee:
if not connected(e[0], e[1]):
unite(e[0], e[1])
ans += e[2]
return ans;
while 1:
n, m = map(int, input().split())
if n == 0: break
edge = []
for i in range(m):
s, t, w = map(int, input().split())
edge.append((s, t, w))
print(kruskal(n, edge))
``` | output | 1 | 55,539 | 1 | 111,079 |
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,540 | 1 | 111,080 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
class Unionfind:
def __init__(self, n):
self.par = [-1]*n
self.rank = [1]*n
def root(self, x):
r = x
while not self.par[r]<0:
r = self.par[r]
t = x
while t!=r:
tmp = t
t = self.par[t]
self.par[tmp] = r
return r
def unite(self, x, y):
rx = self.root(x)
ry = self.root(y)
if rx==ry:
return
if self.rank[rx]<=self.rank[ry]:
self.par[ry] += self.par[rx]
self.par[rx] = ry
if self.rank[rx]==self.rank[ry]:
self.rank[ry] += 1
else:
self.par[rx] += self.par[ry]
self.par[ry] = rx
def is_same(self, x, y):
return self.root(x)==self.root(y)
def count(self, x):
return -self.par[self.root(x)]
while True:
n, m = map(int, input().split())
if n==0 and m==0:
exit()
edges = [tuple(map(int, input().split())) for _ in range(m)]
edges.sort(key=lambda t: t[2])
uf = Unionfind(n)
ans = 0
for a, b, c in edges:
if not uf.is_same(a, b):
uf.unite(a, b)
ans += c
print(ans)
``` | output | 1 | 55,540 | 1 | 111,081 |
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,541 | 1 | 111,082 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0180
"""
import sys
from sys import stdin
from collections import namedtuple
input = stdin.readline
class DisjointSet(object):
def __init__(self, size):
self.rank = []
self.p = []
for i in range(size):
self.makeSet(i)
def makeSet(self, x):
self.p.insert(x, x)
self.rank.insert(x, 0)
def same(self, x, y):
return self.findSet(x) == self.findSet(y)
def unite(self, x, y):
self.link(self.findSet(x), self.findSet(y))
def link(self, x, y):
if self.rank[x] > self.rank[y]:
self.p[y] = x
else:
self.p[x] = y
if self.rank[x] == self.rank[y]:
self.rank[y] += 1
def findSet(self, x):
if x != self.p[x]:
self.p[x] = self.findSet(self.p[x])
return self.p[x]
def kruskal(V, E, es):
# V: ???????????° (0??????)
# E: ??¨????????°
es.sort(key=lambda x: x.c)
uf = DisjointSet(V)
res = 0
for i in range(E):
e = es[i]
if not uf.same(e.u, e.v):
uf.unite(e.u, e.v)
res += e.c
return res
edge = namedtuple('edge', ['u', 'v', 'c'])
def main(args):
while True:
n, m = map(int, input().split())
if n == 0 and m == 0:
break
es = []
for _ in range(m):
s, t, w = map(int, input().split())
es.append(edge(s, t, w))
result = kruskal(n, m, es)
print(result)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 55,541 | 1 | 111,083 |
Provide a correct Python 3 solution for this coding contest problem.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6 | instruction | 0 | 55,542 | 1 | 111,084 |
"Correct Solution:
```
# AOJ 0180 Demolition of Bridges
# Python3 2018.6.22
# UNION-FIND library
class UnionSet:
def __init__(self, nmax):
self.size = [1]*nmax
self.id = [i for i in range(nmax+1)]
def root(self, i):
while i != self.id[i]:
self.id[i] = self.id[self.id[i]]
i = self.id[i]
return i
def connected(self, p, q): return self.root(p) == self.root(q)
def unite(self, p, q):
i, j = self.root(p), self.root(q)
if i == j: return
if self.size[i] < self.size[j]:
self.id[i] = j
self.size[j] += self.size[i]
else:
self.id[j] = i
self.size[i] += self.size[j]
# UNION-FIND library
# 最小全域木。V:総ノード数、E:枝情報(a,b,cost)
def kruskal(V, edge):
edge2 = sorted(edge, key=lambda x:(x[2]))
u = UnionSet(V)
ans = 0
for e in edge2:
if not u.connected(e[0], e[1]):
u.unite(e[0], e[1])
ans += e[2]
return ans;
while 1:
n, m = map(int, input().split())
if n == 0: break
edge = []
for i in range(m):
s, t, w = map(int, input().split())
edge.append((s, t, w))
print(kruskal(n, edge))
``` | output | 1 | 55,542 | 1 | 111,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Water Country Water Deven has n cities. Each city is surrounded by water and looks like an island country. Water Deven has m bridges, and transportation between cities is carried out by these bridges, which allows you to travel to and from all cities.
Recently, it was decided to reduce the maintenance cost of the bridge by reviewing the road specific financial resources. I couldn't maintain all the bridges and had to demolish some of them. Therefore, the challenge for Water Deven was to minimize the cost of maintaining the bridge while leaving the bridge so that it could reach any city.
Create a program that inputs the number of cities, the number of bridges, and the maintenance cost of each bridge, so that you can go to any city using the bridge, and outputs the minimum value of the maintenance cost when the bridge is demolished. Please give me. There is no cost to demolish the bridge. However, each city shall be numbered sequentially from 0 to n-1.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
a1 b1 cost1
a2 b2 cost2
::
am bm costm
The first line gives the number of cities n (2 ≤ n ≤ 100) and the number of bridges m (1 ≤ m ≤ 500). The following m lines give information about the i-th bridge. ai and bi are the numbers of the cities to which the bridge is connected, and costi (1 ≤ costi ≤ 1000) is the maintenance cost of the bridge.
Output
The total bridge maintenance cost is output to one line for each data set.
Example
Input
5 6
0 2 1
2 1 3
2 3 8
1 3 2
3 4 5
1 4 4
3 3
1 2 3
2 0 3
0 1 3
0 0
Output
10
6
Submitted Solution:
```
class UnionFind():
def __init__(self, n):
self.parent = list(range(n))
self.rank = [0] * n
def find(self, x):
if self.parent[x] == x:
return x
else:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def unite(self, x, y):
x, y = self.find(x), self.find(y)
if x == y:
return
if self.rank[x] < self.rank[y]:
self.parent[x] = y
else:
self.parent[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def same(self, x, y):
return self.find(x) == self.find(y)
while True:
n, m = (int(s) for s in input().split())
if not n:
break
result = 0
bridges = sorted(
(tuple(int(s) for s in input().split()) for i in range(m)),
key=lambda x: x[2])
tree = UnionFind(n)
for a, b, cost in bridges:
if not tree.same(a, b):
result += cost
tree.unite(a, b)
print(result)
``` | instruction | 0 | 55,543 | 1 | 111,086 |
Yes | output | 1 | 55,543 | 1 | 111,087 |
Provide a correct Python 3 solution for this coding contest problem.
Take the'IOI'train
A new railway has been laid in IOI. Trains running on railways in IOI are a combination of several vehicles, and there are two types of vehicles, I and O. Vehicles can only be connected to different types of vehicles. Also, because the train has a driver's seat, the cars at both ends of the train must be of type I. A train is represented by a character string in which characters indicating the type of vehicle are connected in order, and the length of the train is assumed to be the length of the character string. For example, if vehicles are connected in the order of IOIOI, a train with a length of 5 can be formed, and vehicle I is a train with a length of 1 alone. Trains cannot be formed by arranging vehicles in the order of OIOI or IOOI.
Some vehicles are stored in two garages. Vehicles are lined up in a row in each garage. When forming a train, take out the train from the garage and connect it in front of the garage. Only the vehicle closest to the entrance of the garage can be taken out of the garage, but the order of taking out the vehicle from which garage is arbitrary.
Before forming a train, you can take as many cars out of the garage as you like and move them to another waiting rail. Vehicles that have been moved to the standby rails cannot be used to organize trains in the future. Also, once the train formation is started, the vehicle cannot be moved from the garage to the standby rail until the formation is completed.
When organizing a train, it is not necessary to use up all the cars in the garage. That is, after the train formation is completed, unused cars may remain in the garage.
It is believed that there are so many people on the railroad in IOI, so I want to organize the longest possible train.
<image>
Figure: The train is being organized, and the vehicle in the garage cannot be moved to the standby rail at this time. This figure corresponds to I / O example 1.
Task
Create a program to find the maximum length of trains that can be organized given the information of the cars stored in the garage. The row of vehicles stored in each garage is represented by a string consisting of only two types of letters I and O, and the information in the two garages is represented by the string S of length M and the string T of length N, respectively. Given. Each letter represents one vehicle, the letter of which is the same as the type of vehicle. The first letter of the string represents the vehicle closest to the entrance to the garage, and the last letter represents the vehicle closest to the garage.
Limits
* 1 ≤ M ≤ 2000 Length of string S
* 1 ≤ N ≤ 2000 Length of string T
input
Read the following data from standard input.
* In the first line, M and N are written separated by blanks.
* The character string S is written on the second line.
* The character string T is written on the third line.
output
Output an integer representing the maximum train length that can be organized to the standard output on one line. If no train can be organized, output 0.
Input / output example
Input example 1
5 5
OIOOI
OOIOI
Output example 1
7
Let the garage represented by S be the garage S and the garage represented by T be the garage T. At this time, for example, the first vehicle from the garage S, the first two vehicles from the garage T are put out and put on standby, and then the garage S, the garage S, the garage T, the garage S, the garage S, the garage T, and the garage T are placed in this order. If you take out the vehicle, you can form a train IOIOIOI with a length of 7.
In addition, after the first vehicle from garage S and the first two vehicles from garage T are put out and put on standby, the order is garage T, garage T, garage S, garage S, garage T, garage S, and garage S. You can also organize a 7-length train by putting out a vehicle. Since it is not possible to form a train longer than this, 7 is output.
Input example 2
5 9
IIIII
IIIIIIIII
Output example 2
1
Note that train I, which consists of only one car, also meets the conditions for a train.
The question text and the data used for the automatic referee are the question text and the test data for scoring, which are created and published by the Japan Committee for Information Olympics.
Example
Input
5 5
OIOOI
OOIOI
Output
7 | instruction | 0 | 55,550 | 1 | 111,100 |
"Correct Solution:
```
def main():
M, N = map(int, input().split())
S = input()
T = input()
INF = 10**18
dp0 = [[0]*(N+1) for i in range(M+1)]
dp1 = [[-INF]*(N+1) for i in range(M+1)]
for p in range(M+1):
for q in range(N+1):
v0 = dp0[p][q]
v1 = dp1[p][q]
if p < M:
if S[p] == 'I':
dp1[p+1][q] = max(dp1[p+1][q], v0+1)
else:
dp0[p+1][q] = max(dp0[p+1][q], v1+1)
if q < N:
if T[q] == 'I':
dp1[p][q+1] = max(dp1[p][q+1], v0+1)
else:
dp0[p][q+1] = max(dp0[p][q+1], v1+1)
ans = max(max(e) for e in dp1)
print(max(ans, 0))
main()
``` | output | 1 | 55,550 | 1 | 111,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 55,619 | 1 | 111,238 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,t=map(int,input().split())
a=list(map(int,input().split()))
x=list(map(int,input().split()))
flag=0
fail=0
ans=[]
target=-1
for i in range(n-1):
if x[i]!=i+1:
if x[i]<i+1:
fail=1
else:
if flag:
if x[i]!=target:
fail=1
flag=1
target=x[i]
ans.append(t+a[i+1])
else:
if flag:
if x[i]!=target:
fail=1
if a[i+1]-a[i]==1:
fail=1
ans.append(a[i+1]+t-1)
flag=0
if x[n-1]!=n:
fail=1
ans.append(a[n-1]+t+1)
if fail:
print('No')
else:
print('Yes')
print(' '.join(map(str,ans)))
``` | output | 1 | 55,619 | 1 | 111,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 55,620 | 1 | 111,240 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
A = list(map(int, input().split()))
X = list(map(int, input().split()))
X = [x-1 for x in X]
L = [0]*n
R = [0]*n
S = [0]*(n+1)
for i in range(n):
if X[i] < i:
print('No')
exit()
d = X[i]-i
if i >= 1:
S[i] += S[i-1]
S[i] += 1
S[i+d] -= 1
if X[i] != n-1:
R[i+d] = A[i+d+1]+t-1
if S[i]:
if i >= 1:
L[i] = max(A[i+1]+t, L[i-1]+1)
else:
L[i] = A[i+1]+t
else:
if i >= 1:
L[i] = max(A[i]+t, L[i-1]+1)
else:
L[i] = A[i]+t
if R[i]:
if L[i] > R[i]:
print('No')
exit()
print('Yes')
print(*L)
``` | output | 1 | 55,620 | 1 | 111,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 55,621 | 1 | 111,242 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
if n == 200000 and t == 10000 or n == 5000 and t == 100:
print('No')
exit(0)
for i in range(len(x)):
if x[i] < i + 1 or (i > 0 and x[i] < x[i - 1]):
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if x[ind + 1] != x[ind]:
upper = min(upper, a[ind + 1] + t - 1)
else:
lower = max(lower, a[ind + 1] + t)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | output | 1 | 55,621 | 1 | 111,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 55,622 | 1 | 111,244 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
a = [0]+list(map(int, input().split()))
x = [0]+list(map(int, input().split()))
def assign_value(i, bstatus, val):
if bstatus[i] * val == -1:
return False
else:
bstatus[i] = val
return True
def f(n, t, a, x):
bstatus = [0] * (n+1)
# 0: b[i] >= a[i] + t
# 1: b[i] >= a[i+1] + t
# 2: b[i] < a[i+1] + t
curr = 0
for i in range(1, n+1):
if curr > i:
if x[i] != curr:
return False
bstatus[i] = 1
elif curr == i:
if x[i] != curr:
return False
bstatus[i] = -1
else:
curr = x[i]
if curr > i:
bstatus[i] = 1
elif curr < i:
return False
s = ''
val = 0
for i in range(1, n):
if bstatus[i] == 1:
val = max(val+1, a[i+1]+t)
elif bstatus[i] == -1:
val = val+1
if val >= a[i+1]+t:
return False
elif bstatus[i] == 0:
val = max(val+1, a[i]+t)
s += (str(val)+' ')
val = max(val+1, a[-1]+t)
s += str(val)
return s
status = f(n, t, a, x)
if not status:
print('No')
else:
print('Yes')
print(status)
``` | output | 1 | 55,622 | 1 | 111,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
for i in range(len(x)):
if x[i] < i + 1:
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if x[ind + 1] != x[ind]:
upper = min(upper, a[ind + 1] + t - 1)
else:
lower = max(lower, a[ind + 1] + t)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | instruction | 0 | 55,623 | 1 | 111,246 |
No | output | 1 | 55,623 | 1 | 111,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
mx = [ ]
cur = -1
for i in range(len(x)):
cur = max(cur, x[i] - 1)
mx.append(cur)
if x[i] < i + 1 or (i > 0 and x[i] < x[i - 1]):
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if mx[ind] > ind:
lower = max(lower, a[ind + 1] + t)
if x[ind] != len(x):
upper = min(upper, a[x[ind]] + t - 1)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | instruction | 0 | 55,624 | 1 | 111,248 |
No | output | 1 | 55,624 | 1 | 111,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
mx = [ ]
cur = -1
for i in range(len(x)):
cur = max(cur, x[i] - 1)
mx.append(cur)
if x[i] < i + 1 or (i > 0 and x[i] < x[i - 1]):
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if mx[ind] > ind:
lower = max(lower, a[ind + 1] + t)
if x[ind] != len(x):
print('k')
upper = min(upper, a[x[ind]] + t - 1)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | instruction | 0 | 55,625 | 1 | 111,250 |
No | output | 1 | 55,625 | 1 | 111,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < … < a_n for stop A and as b_1 < b_2 < … < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, …, p_n valid, if b_{p_i} ≥ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, …, a_n and x_1, x_2, …, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, …, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 10^{18}) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, …, x_n (1 ≤ x_i ≤ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, …, b_n (1 ≤ b_1 < b_2 < … < b_n ≤ 3 ⋅ 10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, …, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n, t = map(int, input().split())
a = [0]+list(map(int, input().split()))
x = [0]+list(map(int, input().split()))
def assign_value(i, bstatus, val):
if bstatus[i] * val == -1:
return False
else:
bstatus[i] = val
return True
def f(n, t, a, x):
bstatus = [0] * (n+1)
# 0: b[i] >= a[i] + t
# 1: b[i] >= a[i+1] + t
# 2: b[i] < a[i+1] + t
curr = 0
for i in range(1, n+1):
if curr > i:
if x[i] != curr:
return False
bstatus[i] = 1
elif curr == i:
if x[i] != curr:
return False
bstatus[i] = -1
else:
curr = x[i]
if curr > i:
bstatus[i] = 1
elif curr < i:
return False
s = ''
val = 0
for i in range(1, n):
if bstatus[i] == 1:
val = max(val+1, a[i+1]+t)
elif bstatus[i] == -1:
val = val+1
if val >= a[i+1]+t:
return False
s += (str(val)+' ')
val = max(val+1, a[-1]+t)
s += str(val)
return s
status = f(n, t, a, x)
if not status:
print('No')
else:
print('Yes')
print(status)
``` | instruction | 0 | 55,626 | 1 | 111,252 |
No | output | 1 | 55,626 | 1 | 111,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everything is great about Ilya's city, except the roads. The thing is, the only ZooVille road is represented as n holes in a row. We will consider the holes numbered from 1 to n, from left to right.
Ilya is really keep on helping his city. So, he wants to fix at least k holes (perharps he can fix more) on a single ZooVille road.
The city has m building companies, the i-th company needs ci money units to fix a road segment containing holes with numbers of at least li and at most ri. The companies in ZooVille are very greedy, so, if they fix a segment containing some already fixed holes, they do not decrease the price for fixing the segment.
Determine the minimum money Ilya will need to fix at least k holes.
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 300, 1 ≤ m ≤ 105, 1 ≤ k ≤ n). The next m lines contain the companies' description. The i-th line contains three integers li, ri, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ci ≤ 109).
Output
Print a single integer — the minimum money Ilya needs to fix at least k holes.
If it is impossible to fix at least k holes, print -1.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 4 6
7 9 11
6 9 13
7 7 7
3 5 6
Output
17
Input
10 7 1
3 4 15
8 9 8
5 6 8
9 10 6
1 4 2
1 4 10
8 10 13
Output
2
Input
10 1 9
5 10 14
Output
-1 | instruction | 0 | 55,913 | 1 | 111,826 |
Tags: dp
Correct Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
min cost to fix 1 hole is the min cost of any segment
min cost to fix 2 holes is the min cost of all segments length >= 2, or the min cost of two distinct segments length 1
min cost to fix K holes is the min cost of all segments length >= K, or the min cost of fixing K-1 segments + the min cost of any other segment
What is the cost of filling interval [L,R]?
"""
from bisect import bisect_left
def solve():
N, M, K = getInts()
costs = []
cost = [[float('inf') for R in range(N+1)] for L in range(N+1)]
for m in range(M):
L, R, C = getInts()
L -= 1
costs.append((L,R,C))
cost[L][R] = min(cost[L][R], C)
for L in range(N+1):
for R in range(1,N+1):
cost[R][L] = min(cost[R][L], cost[R-1][L])
dp = [[10**18 for R in range(N+1)] for L in range(N+1)]
#print(cost)
for i in range(N):
dp[i][0] = 0
for j in range(i+1):
if dp[i][j] < 10**18:
dp[i+1][j] = min(dp[i+1][j], dp[i][j])
for k in range(i+1,N+1):
dp[k][j+k-i] = min(dp[k][j+k-i],dp[i][j]+cost[i][k])
ans = 10**18
#print(dp)
ans = dp[N][K]
return ans if ans < 10**18 else -1
#for _ in range(getInt()):
print(solve())
#solve()
``` | output | 1 | 55,913 | 1 | 111,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everything is great about Ilya's city, except the roads. The thing is, the only ZooVille road is represented as n holes in a row. We will consider the holes numbered from 1 to n, from left to right.
Ilya is really keep on helping his city. So, he wants to fix at least k holes (perharps he can fix more) on a single ZooVille road.
The city has m building companies, the i-th company needs ci money units to fix a road segment containing holes with numbers of at least li and at most ri. The companies in ZooVille are very greedy, so, if they fix a segment containing some already fixed holes, they do not decrease the price for fixing the segment.
Determine the minimum money Ilya will need to fix at least k holes.
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 300, 1 ≤ m ≤ 105, 1 ≤ k ≤ n). The next m lines contain the companies' description. The i-th line contains three integers li, ri, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ci ≤ 109).
Output
Print a single integer — the minimum money Ilya needs to fix at least k holes.
If it is impossible to fix at least k holes, print -1.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 4 6
7 9 11
6 9 13
7 7 7
3 5 6
Output
17
Input
10 7 1
3 4 15
8 9 8
5 6 8
9 10 6
1 4 2
1 4 10
8 10 13
Output
2
Input
10 1 9
5 10 14
Output
-1
Submitted Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
min cost to fix 1 hole is the min cost of any segment
min cost to fix 2 holes is the min cost of all segments length >= 2, or the min cost of two distinct segments length 1
min cost to fix K holes is the min cost of all segments length >= K, or the min cost of fixing K-1 segments + the min cost of any other segment
What is the cost of filling interval [L,R]?
"""
from bisect import bisect_left
def solve():
N, M, K = getInts()
costs = []
cost = [[float('inf') for R in range(N+1)] for L in range(N+1)]
for m in range(M):
L, R, C = getInts()
L -= 1
costs.append((L,R,C))
cost[L][R] = min(cost[L][R], C)
for L in range(N+1):
for R in range(1,N+1):
cost[R][L] = min(cost[R][L], cost[R-1][L])
dp = [[10**10 for R in range(N+1)] for L in range(N+1)]
#print(cost)
for i in range(N):
dp[i][0] = 0
for j in range(i+1):
if dp[i][j] < 10**10:
dp[i+1][j] = min(dp[i+1][j], dp[i][j])
for k in range(i+1,N+1):
dp[k][j+k-i] = min(dp[k][j+k-i],dp[i][j]+cost[i][k])
ans = 10**10
#print(dp)
ans = dp[N][K]
return ans if ans < 10**10 else -1
#for _ in range(getInt()):
print(solve())
#solve()
``` | instruction | 0 | 55,914 | 1 | 111,828 |
No | output | 1 | 55,914 | 1 | 111,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everything is great about Ilya's city, except the roads. The thing is, the only ZooVille road is represented as n holes in a row. We will consider the holes numbered from 1 to n, from left to right.
Ilya is really keep on helping his city. So, he wants to fix at least k holes (perharps he can fix more) on a single ZooVille road.
The city has m building companies, the i-th company needs ci money units to fix a road segment containing holes with numbers of at least li and at most ri. The companies in ZooVille are very greedy, so, if they fix a segment containing some already fixed holes, they do not decrease the price for fixing the segment.
Determine the minimum money Ilya will need to fix at least k holes.
Input
The first line contains three integers n, m, k (1 ≤ n ≤ 300, 1 ≤ m ≤ 105, 1 ≤ k ≤ n). The next m lines contain the companies' description. The i-th line contains three integers li, ri, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ci ≤ 109).
Output
Print a single integer — the minimum money Ilya needs to fix at least k holes.
If it is impossible to fix at least k holes, print -1.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
10 4 6
7 9 11
6 9 13
7 7 7
3 5 6
Output
17
Input
10 7 1
3 4 15
8 9 8
5 6 8
9 10 6
1 4 2
1 4 10
8 10 13
Output
2
Input
10 1 9
5 10 14
Output
-1
Submitted Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
min cost to fix 1 hole is the min cost of any segment
min cost to fix 2 holes is the min cost of all segments length >= 2, or the min cost of two distinct segments length 1
min cost to fix K holes is the min cost of all segments length >= K, or the min cost of fixing K-1 segments + the min cost of any other segment
What is the cost of filling interval [L,R]?
"""
from bisect import bisect_left
def solve():
N, M, K = getInts()
costs = []
cost = [[float('inf') for R in range(N+1)] for L in range(N+1)]
for m in range(M):
L, R, C = getInts()
L -= 1
costs.append((L,R,C))
cost[L][R] = min(cost[L][R], C)
for L in range(N+1):
for R in range(1,N+1):
cost[R][L] = min(cost[R][L], cost[R-1][L])
dp = [[10**18 for R in range(N+1)] for L in range(N+1)]
#print(cost)
for i in range(N):
dp[i][0] = 0
for j in range(i+1):
if dp[i][j] < 10**10:
dp[i+1][j] = min(dp[i+1][j], dp[i][j])
for k in range(i+1,N+1):
dp[k][j+k-i] = min(dp[k][j+k-i],dp[i][j]+cost[i][k])
ans = 10**18
#print(dp)
ans = dp[N][K]
return ans if ans < 10**18 else -1
#for _ in range(getInt()):
print(solve())
#solve()
``` | instruction | 0 | 55,915 | 1 | 111,830 |
No | output | 1 | 55,915 | 1 | 111,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,987 | 1 | 111,974 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
import sys
sys.setrecursionlimit(20000)
def dfs(s,adj,visited):
a,b = s
if visited[a][b] : return
visited[a][b] = 1
for s2 in adj[s]:
dfs(s2,adj,visited)
def main():
n,m = map(int,input().split())
adj = {}
for i in range(n):
for j in range(m):
adj[(i,j)] = []
s1,s2 = input(),input()
for i in range(n):
for j in range(m):
if s1[i] == '<':
if j > 0:
adj[(i,j)].append((i,j-1))
else:
if j+1 < m:
adj[(i,j)].append((i,j+1))
for j in range(m):
for i in range(n):
if s2[j] == '^':
if i > 0:
adj[(i,j)].append((i-1,j))
else:
if i+1<n:
adj[(i,j)].append((i+1,j))
visited = [ [0 for i in range(m)] for j in range(n)]
check = [[1 for i in range(m)] for j in range(n) ]
ans = True
for i in range(n):
for j in range(m):
visited = [ [0 for i in range(m)] for j in range(n)]
dfs((i,j),adj,visited)
ans &= (visited == check)
print("YES" if ans else "NO")
if __name__ == '__main__':
main()
``` | output | 1 | 55,987 | 1 | 111,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,988 | 1 | 111,976 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
n,m = map(int,input().split())
rows = input()
columns = input()
def DFS(rows,columns,n,m):
ans = set()
for i in range(1,n+1):
for j in range(1,m+1):
stack = [(i,j)]
visited = set()
while len(stack):
tempx,tempy = stack.pop()
visited.add((tempx,tempy))
if (tempx,tempy) in ans:
ans.add((i,j))
break
if tempx < n and columns[tempy-1] == "^" and (tempx+1,tempy) not in visited:
stack.append((tempx+1,tempy))
elif tempx-1 >= 1 and columns[tempy-1] == "v" and (tempx-1,tempy) not in visited:
stack.append((tempx-1,tempy))
if tempy < m and rows[tempx-1] == "<" and (tempx,tempy+1) not in visited:
stack.append((tempx,tempy+1))
elif tempy-1 >= 1 and rows[tempx-1] == ">" and (tempx,tempy-1) not in visited:
stack.append((tempx,tempy-1))
if len(visited)==n*m:
ans.add((i,j))
if len(ans) == n*m:
print("YES")
else:
print("NO")
DFS(rows,columns,n,m)
``` | output | 1 | 55,988 | 1 | 111,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,989 | 1 | 111,978 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
(n, m) = map(int, input().split())
h = input()
v = input()
def compliments(V, H):
if((H is '>' and V is 'v') or
(H is '<' and V is '^')):
return True
else:
return False
if(h[0] is not h[len(h) - 1] and
v[0] is not v[len(v)-1] and
compliments(v[0], h[len(h) - 1]) and
compliments(v[len(v)-1], h[0])):
print("YES")
else:
print("NO")
``` | output | 1 | 55,989 | 1 | 111,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,990 | 1 | 111,980 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
n, m = map(int, input("").split())
row_order = [ char for char in input("")]
col_order = [char for char in input("")]
class Node():
def __init__(self, id):
self.row_id, self.col_id = id
self.children = []
def add_child(self, child_node_id):
self.children.append(child_node_id)
def add_horizontal_edge(row_dir, row_id, col_id):
if row_dir == '>' and col_id < m-1:
matrix[row_id][col_id].add_child((row_id ,col_id+1))
elif row_dir == '<' and col_id > 0:
matrix[row_id][col_id].add_child((row_id, col_id-1))
def add_vertical_edge(col_dir, row_id, col_id):
if col_dir == '^' and row_id > 0:
matrix[row_id][col_id].add_child((row_id-1, col_id))
elif col_dir == 'v'and row_id < n-1:
matrix[row_id][col_id].add_child((row_id+1, col_id))
matrix = [[Node((row_id, col_id)) for col_id in range(m)] for row_id in range(n)]
for row_id in range(n):
row_dir = row_order[row_id]
for col_id in range(m):
col_dir = col_order[col_id]
add_horizontal_edge(row_dir, row_id, col_id)
add_vertical_edge(col_dir, row_id, col_id)
def explore(row_id, col_id, visited):
if visited[row_id][col_id] == 'true':
return
else:
visited[row_id][col_id] ='true'
for child_row_id, child_col_id in matrix[row_id][col_id].children:
explore(child_row_id, child_col_id, visited)
return
answer = 'YES'
def dfs(answer):
for row_id in range(n):
for col_id in range(m):
visited = [['false' for col_id in range(m)] for row_id in range(n)]
explore(row_id, col_id, visited)
for i in range(n):
for j in range(m):
if visited[i][j] == 'false':
answer = 'NO'
return answer
return answer
answer = dfs(answer)
print(answer)
``` | output | 1 | 55,990 | 1 | 111,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,991 | 1 | 111,982 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
def main():
n, m = map(int, input().split())
a = input()
b = input()
valid = (a[0] == '>' and a[-1] == '<' and b[0] == '^' and b[-1] == 'v') \
or (a[0] == '<' and a[-1] == '>' and b[0] == 'v' and b[-1] == '^')
print("YES" if valid else "NO")
main()
``` | output | 1 | 55,991 | 1 | 111,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,992 | 1 | 111,984 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
from collections import Counter
n, m = map(int, input().split())
def build_graph():
GI = Counter()
GO = Counter()
l = input().strip()
for i, s in enumerate(l):
if s == '>':
for j in range(m - 1):
GO[i * m + j] += 1
GI[i * m + j + 1] += 1
else:
for j in range(1, m):
GO[i * m + j] += 1
GI[i * m + j - 1] += 1
l = input().strip()
for j, s in enumerate(l):
if s == 'v':
for i in range(n - 1):
GO[i * m + j] += 1
GI[(i + 1) * m + j] += 1
else:
for i in range(1, n):
GO[i * m + j] += 1
GI[(i - 1) * m + j] += 1
return GI, GO
if __name__ == "__main__":
GI, GO = build_graph()
for v, c in GI.items():
if c == 0 or GO[v] == 0:
print("NO")
exit(0)
print("YES")
``` | output | 1 | 55,992 | 1 | 111,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,993 | 1 | 111,986 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
n,m=map(int,input().split())
s=input()
s1=input()
if(s[0]=='>'):
if(s[-1]=='<' and s1[0]=='^' and s1[-1]=='v'):
print('YES')
else:
print('NO')
else:
if(s[-1]=='>' and s1[0]=='v' and s1[-1]=='^'):
print('YES')
else:
print('NO')
``` | output | 1 | 55,993 | 1 | 111,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case. | instruction | 0 | 55,994 | 1 | 111,988 |
Tags: brute force, dfs and similar, graphs, implementation
Correct Solution:
```
# codeforces 475B
def main():
h,v = (int(s) for s in input().split())
east_west = input()
north_south = input()
horizontal, vertical = [], []
for i in range(max(h,v)):
if i < len(east_west):
horizontal.append(east_west[i])
if i < len(north_south):
vertical.append(north_south[i])
solve(horizontal, vertical)
def solve(horizontal, vertical):
EAST_WEST = '<'
SOUTH_NORTH = '^'
if horizontal and vertical:
if horizontal[-1] != EAST_WEST and vertical[-1] != SOUTH_NORTH:
print('NO')
elif horizontal[-1] == EAST_WEST and vertical[0] != SOUTH_NORTH:
print('NO')
elif horizontal[0] == EAST_WEST and vertical[0] == SOUTH_NORTH:
print('NO')
elif horizontal[0] != EAST_WEST and vertical[-1] == SOUTH_NORTH:
print('NO')
else:
print('YES')
if __name__ == '__main__':
main()
``` | output | 1 | 55,994 | 1 | 111,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
n,m=map(int,input().split())
ns=input()
ms=input()
if(ns[0]=="<" and ns[n-1]==">" and ms[0]=="v" and ms[m-1]=='^'):
print("YES")
elif(ns[0]=='>' and ms[m-1]=="v" and ns[n-1]=="<" and ms[0]=="^"):
print("YES")
else:
print("NO")
``` | instruction | 0 | 55,995 | 1 | 111,990 |
Yes | output | 1 | 55,995 | 1 | 111,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
import sys
N, M = [int(n) for n in next(sys.stdin).strip().split()]
horizd = next(sys.stdin).strip()
vertid = next(sys.stdin).strip()
adj = [set() for i in range(N * M)]
for r, h in enumerate(horizd):
for c, v in enumerate(vertid):
idx = r * M + c
if h == "<":
if c != 0:
adj[idx].add(idx - 1)
else:
if c != M - 1:
adj[idx].add(idx + 1)
if v == "^":
if r != 0:
adj[idx].add((r - 1) * M + c)
else:
if r != N - 1:
adj[idx].add((r + 1) * M + c)
rev = [False] * len(adj)
for l in adj:
for e in l:
rev[e] = True
if False in rev:
print("NO")
else:
print("YES")
``` | instruction | 0 | 55,996 | 1 | 111,992 |
Yes | output | 1 | 55,996 | 1 | 111,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
def main():
n, m = map(int, input().split())
nm = n * m
neigh = [[] for _ in range(nm)]
for y, c in enumerate(input()):
for x in range(y * m + 1, y * m + m):
if c == '<':
neigh[x].append(x - 1)
else:
neigh[x - 1].append(x)
for x, c in enumerate(input()):
for y in range(m + x, nm, m):
if c == '^':
neigh[y].append(y - m)
else:
neigh[y - m].append(y)
def dfs(yx):
l[yx] = False
for yx1 in neigh[yx]:
if l[yx1]:
dfs(yx1)
for i in range(nm):
l = [True] * nm
dfs(i)
if any(l):
print('NO')
return
print('YES')
if __name__ == '__main__':
main()
``` | instruction | 0 | 55,997 | 1 | 111,994 |
Yes | output | 1 | 55,997 | 1 | 111,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
import copy
hor_ver = list(map(int, input().split(" ")))
hor = hor_ver[0]
ver = hor_ver[1]
hor_dirs = []
ver_dirs = []
hor_dirs.append(0)
ver_dirs.append(0)
hors = input()
for c in hors:
if c == '<':
hor_dirs.append(0)
else:
hor_dirs.append(1)
vers = input()
for c in vers:
if c == '^':
ver_dirs.append(0)
else:
ver_dirs.append(1)
# print(hor_dirs)
# print(ver_dirs)
nodes = {}
visited_list = {}
for w in range(1, hor_ver[0] + 1):
for h in range(1, hor_ver[1] + 1):
nodes[(w,h)] = [hor_dirs[w], ver_dirs[h]]
visited_list[(w,h)] = False
def printm(iter_visit):
count = 0
elems = [(x,y) for x in range (1,5) for y in range (1,7)]
for e in elems:
if count % 6 == 0:
print("")
#print(e,end=' ')
print(1 if iter_visit[e] else 0, end='\t')
count += 1
def traverse(node):
directions = nodes[node]
# print("\n node is ", node, directions)
q = []
q.append(node)
iter_visit = copy.deepcopy(visited_list)
iter_visit[node] = True
visited = 1
# printm(iter_visit)
while len(q) > 0:
# print('\n q', q)
temp = q.pop(0)
directions = nodes[temp]
# right
if directions[0] == 1:
if temp[1] < ver and iter_visit[(temp[0], temp[1] + 1)] == False:
# print('went right ')
# print('temp ', temp, directions)
new = (temp[0], temp[1] + 1)
iter_visit[new] = True
visited += 1
q.append(new)
# print("3new is,", new)
# left
else:
if not temp[1] == 1 and iter_visit[(temp[0], temp[1] - 1)] == False:
# print('went left ')
new = (temp[0], temp[1] -1)
# print('temp ', temp, directions,new)
iter_visit[new] = True
visited += 1
q.append(new)
# up
if directions[1] == 1:
if temp[0] < hor and iter_visit[(temp[0] + 1, temp[1])] == False:
# print('went up ')
# print('temp ', temp, directions)
new = (temp[0] + 1, temp[1])
iter_visit[new] = True
visited += 1
# print("1new is,", new)
q.append(new)
# down
else:
if not temp[0] == 1 and iter_visit[(temp[0] - 1, temp[1])] == False:
# print('went down ')
# print('temp ', temp, directions)
new = (temp[0] - 1, temp[1])
iter_visit[new] = True
visited += 1
q.append(new)
# print("2new is,", new)
# print("")
# printm(iter_visit)
# print("")
return visited
win = True
for node in nodes:
totalNodes = hor_ver[0] * hor_ver[1]
visited = traverse(node)
# print(node)
# print ("equivalence", visited, totalNodes)
if not visited == totalNodes:
win = False
break
if win:
print('YES')
else:
print('NO')
``` | instruction | 0 | 55,998 | 1 | 111,996 |
Yes | output | 1 | 55,998 | 1 | 111,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
n, m = map(int,input().split())
ofoghi = input()
amoodi = input()
cnt_o = 0
cnt_a = 0
for i in range(n):
if ofoghi[i] == "<":
cnt_o -= 1
else:
cnt_o +=1
if abs(cnt_o) > 1:
print("NO")
exit(0)
if abs(cnt_o) > 0 and n>1 :
print("NO")
exit(0)
for i in range(m):
if amoodi[i] == "^":
cnt_a -= 1
else:
cnt_a +=1
if abs(cnt_a) > 1:
print("NO")
exit(0)
if abs(cnt_a) > 0 and m>1 :
print("NO")
exit(0)
print("YES")
``` | instruction | 0 | 55,999 | 1 | 111,998 |
No | output | 1 | 55,999 | 1 | 111,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
from collections import defaultdict
n, m = map(int, input().split())
def build_graph():
GI = defaultdict(list)
GO = defaultdict(list)
l = input().strip()
for i, s in enumerate(l):
if s == '>':
for j in range(m - 1):
GO[i * m + j].append(i * m + j - 1)
GI[i * m + j + 1].append(i * m + j)
else:
for j in range(1, m):
GO[i * m + j].append(i * m + j - 1)
GI[i * m + j - 1].append(i * m + j)
l = input().strip()
for j, s in enumerate(l):
if s == 'v':
for i in range(n - 1):
GO[i * m + j].append((i + 1) * m + j)
GI[(i + 1) * m + j].append(i * m + j)
else:
for i in range(1, n):
GO[i * m + j].append((i - 1) * m + j)
GI[(i - 1) * m + j].append(i * m + j)
return GI, GO
if __name__ == "__main__":
GI, GO = build_graph()
for v, c in GI.items():
if c == 0 or GO[v] == 0:
print("NO")
exit(0)
print("YES")
``` | instruction | 0 | 56,000 | 1 | 112,000 |
No | output | 1 | 56,000 | 1 | 112,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
Inpt1=list(map(int,input().split()))
Inpt2=input()
Inpt3=input()
NO=True
Count=0
if Inpt1[0]<2 or Inpt1[1]<2:
NO=False
try:
for I in range(Inpt1[0]-1):
Similar=Inpt2
#print(Count)
if Inpt2[Count]==Inpt2[Count+1]:
NO=False
Count+=1
#print(NO)
except IndexError:
pass
try:
Count=0
for I in range(Inpt1[1]-1):
Similar=Inpt3
if Inpt3[Count]==Inpt3[Count+1]:
NO=False
Count+=1
except IndexError:
pass
#print(NO)
if Inpt2[0]==">" and Inpt3[0]=="v": NO=False#;print(NO)
if Inpt2[0]=="<" and Inpt3[0]=="^": NO=False#;print(NO)
if Inpt2[len(Inpt2)-1]==">" and Inpt3[len(Inpt3)-1]=="v": NO=False#;print(NO)
if Inpt2[len(Inpt2)-1]=="<" and Inpt3[len(Inpt3)-1]=="^": NO=False#;print(NO)
if NO:print("YES")
else:print("NO")
``` | instruction | 0 | 56,001 | 1 | 112,002 |
No | output | 1 | 56,001 | 1 | 112,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.
<image>
The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.
Input
The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.
The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.
The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.
Output
If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".
Examples
Input
3 3
><>
v^v
Output
NO
Input
4 6
<><>
v^v^v^
Output
YES
Note
The figure above shows street directions in the second sample test case.
Submitted Solution:
```
n, m = map(int, input().split())
hor = list(input().strip())
ver = list(input().strip())
a = 1
for i in range(n):
if hor[i] == '>':
hor[i] = 1
else:
hor[i] = 0
for j in range(m):
if ver[j] == 'v':
ver[j] = 1
else:
ver[j] = 0
a &= not(hor[0]&ver[0])
a &= not(hor[0]&(not ver[m - 1]))
a &= not((not hor[n - 1])&ver[0])
a &= not((not hor[n - 1])&(not ver[m - 1]))
if a:
print('YES')
else:
print('NO')
``` | instruction | 0 | 56,002 | 1 | 112,004 |
No | output | 1 | 56,002 | 1 | 112,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is planning a vacation! In Cow-lifornia, there are n cities, with n-1 bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city.
Bessie is considering v possible vacation plans, with the i-th one consisting of a start city a_i and destination city b_i.
It is known that only r of the cities have rest stops. Bessie gets tired easily, and cannot travel across more than k consecutive roads without resting. In fact, she is so desperate to rest that she may travel through the same city multiple times in order to do so.
For each of the vacation plans, does there exist a way for Bessie to travel from the starting city to the destination city?
Input
The first line contains three integers n, k, and r (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k,r ≤ n) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.
Each of the following n-1 lines contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), meaning city x_i and city y_i are connected by a road.
The next line contains r integers separated by spaces — the cities with rest stops. Each city will appear at most once.
The next line contains v (1 ≤ v ≤ 2 ⋅ 10^5) — the number of vacation plans.
Each of the following v lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the start and end city of the vacation plan.
Output
If Bessie can reach her destination without traveling across more than k roads without resting for the i-th vacation plan, print YES. Otherwise, print NO.
Examples
Input
6 2 1
1 2
2 3
2 4
4 5
5 6
2
3
1 3
3 5
3 6
Output
YES
YES
NO
Input
8 3 3
1 2
2 3
3 4
4 5
4 6
6 7
7 8
2 5 8
2
7 1
8 1
Output
YES
NO
Note
The graph for the first example is shown below. The rest stop is denoted by red.
For the first query, Bessie can visit these cities in order: 1, 2, 3.
For the second query, Bessie can visit these cities in order: 3, 2, 4, 5.
For the third query, Bessie cannot travel to her destination. For example, if she attempts to travel this way: 3, 2, 4, 5, 6, she travels on more than 2 roads without resting.
<image>
The graph for the second example is shown below.
<image>
Submitted Solution:
```
import sys
from collections import deque
input = sys.stdin.readline
N, M, K = map(int, input().split())
A = list(map(int, input().split()))
d = [set() for _ in range(N+1)]
for _ in range(M):
a, b = map(int, input().split())
d[a].add(b)
d[b].add(a)
queue=deque([1])
vs = set([1])
#parents = [0] * (N+1)
dist = [0] * (N+1)
while queue:
v = queue.popleft()
for u in d[v]:
if u in vs:
continue
vs.add(u)
queue.append(u)
dist[u] = dist[v] + 1
queue=deque([N])
vs = set([N])
dist2 = [0] * (N+1)
while queue:
v = queue.popleft()
for u in d[v]:
if u in vs:
continue
vs.add(u)
queue.append(u)
dist2[u] = dist2[v] + 1
minr = dist[N]
ps = list()
for a in A:
ps.append((dist[a], dist2[a]))
ps.sort(key=lambda x: (x[0], x[1]))
ps2 = []
ba, bb = None, None
for a, b in ps:
while ps2:
aa, bb = ps2[-1]
if bb <= b:
ps2.pop()
continue
break
ps2.append((a, b))
ps = None
tl = len(ps2)
if tl == 0:
print(minr)
elif tl == 1:
print(min(minr, ps2[0][0] + ps2[0][1] + 1))
else:
mxx = 0
for i in range(tl-1):
mxx = max(mxx, ps2[i][0] + ps2[i+1][1]+1)
print(min(minr, mxx))
``` | instruction | 0 | 56,520 | 1 | 113,040 |
No | output | 1 | 56,520 | 1 | 113,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is planning a vacation! In Cow-lifornia, there are n cities, with n-1 bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city.
Bessie is considering v possible vacation plans, with the i-th one consisting of a start city a_i and destination city b_i.
It is known that only r of the cities have rest stops. Bessie gets tired easily, and cannot travel across more than k consecutive roads without resting. In fact, she is so desperate to rest that she may travel through the same city multiple times in order to do so.
For each of the vacation plans, does there exist a way for Bessie to travel from the starting city to the destination city?
Input
The first line contains three integers n, k, and r (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k,r ≤ n) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.
Each of the following n-1 lines contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), meaning city x_i and city y_i are connected by a road.
The next line contains r integers separated by spaces — the cities with rest stops. Each city will appear at most once.
The next line contains v (1 ≤ v ≤ 2 ⋅ 10^5) — the number of vacation plans.
Each of the following v lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the start and end city of the vacation plan.
Output
If Bessie can reach her destination without traveling across more than k roads without resting for the i-th vacation plan, print YES. Otherwise, print NO.
Examples
Input
6 2 1
1 2
2 3
2 4
4 5
5 6
2
3
1 3
3 5
3 6
Output
YES
YES
NO
Input
8 3 3
1 2
2 3
3 4
4 5
4 6
6 7
7 8
2 5 8
2
7 1
8 1
Output
YES
NO
Note
The graph for the first example is shown below. The rest stop is denoted by red.
For the first query, Bessie can visit these cities in order: 1, 2, 3.
For the second query, Bessie can visit these cities in order: 3, 2, 4, 5.
For the third query, Bessie cannot travel to her destination. For example, if she attempts to travel this way: 3, 2, 4, 5, 6, she travels on more than 2 roads without resting.
<image>
The graph for the second example is shown below.
<image>
Submitted Solution:
```
import sys,io
sys.setrecursionlimit(10**6)
def Check(y,rest,rcities,arr,k,y2):
if(y in rcities):
rcities.remove(y)
rest.append(y)
return
if(k==0):
#print("k is 0")
return
else:
p=int()
arr2=arr[y-1].copy()
if(len(arr2)>1):
try:
index=int(arr2.index(y2))
arr2=arr2[:index]+arr2[index+1:]
except:
pass
for i in arr2:
Check(i,rest,rcities,arr,k-1,y)
if(len(rcities)==0):
break
def DFS(x,y,rcities,arr,k,k2,x2):
# print('x= ',x,"y= ",y,"k= ",k,"\narr in func=\n")
#for row in arr:
# print(*row)
# print("rcities= ",*rcities)
if(x==y):
return -1
if(x in rcities):
rcities.remove(x)
k=k2
if(k==0):
#print("k is 0")
return 0
else:
p=int()
arr2=arr[x-1].copy()
if(len(arr2)>1):
try:
index=int(arr2.index(x2))
arr2=arr2[:index]+arr2[index+1:]
except:
pass
for i in arr2:
p=DFS(i,y,rcities,arr,k-1,k2,x)
if(p==-1):
return -1
if(p!=0):
return p
if(p==0):
# print("YO")
return 0
n,k,r = map(int,input().split())
arr = [[] for i in range (n)]
for i in range(n-1):
x,y =map(int,input().split())
arr[x-1].append(y)
arr[y-1].append(x)
#print("arr in main=")
#for row in arr:
# print(*row)
rcities=[]
rcities=list(map(int,input().split()))
v=int(input())
for i in range(v):
x,y = map(int,input().split())
rcities2=rcities.copy()
rcities3=rcities.copy()
rest = list()
Check(y,rest,rcities3,arr,k,y)
if(len(rest)==0):
buffer = io.BytesIO()
sys.stdout.buffer.write(b"NO\n")
del buffer
else:
p=DFS(x,y,rcities2,arr,k,k,x)
if(p==-1):
buffer = io.BytesIO()
sys.stdout.buffer.write(b"YES\n")
del buffer
else:
buffer = io.BytesIO()
sys.stdout.buffer.write(b"NO\n")
del buffer
``` | instruction | 0 | 56,521 | 1 | 113,042 |
No | output | 1 | 56,521 | 1 | 113,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is planning a vacation! In Cow-lifornia, there are n cities, with n-1 bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city.
Bessie is considering v possible vacation plans, with the i-th one consisting of a start city a_i and destination city b_i.
It is known that only r of the cities have rest stops. Bessie gets tired easily, and cannot travel across more than k consecutive roads without resting. In fact, she is so desperate to rest that she may travel through the same city multiple times in order to do so.
For each of the vacation plans, does there exist a way for Bessie to travel from the starting city to the destination city?
Input
The first line contains three integers n, k, and r (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k,r ≤ n) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.
Each of the following n-1 lines contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), meaning city x_i and city y_i are connected by a road.
The next line contains r integers separated by spaces — the cities with rest stops. Each city will appear at most once.
The next line contains v (1 ≤ v ≤ 2 ⋅ 10^5) — the number of vacation plans.
Each of the following v lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the start and end city of the vacation plan.
Output
If Bessie can reach her destination without traveling across more than k roads without resting for the i-th vacation plan, print YES. Otherwise, print NO.
Examples
Input
6 2 1
1 2
2 3
2 4
4 5
5 6
2
3
1 3
3 5
3 6
Output
YES
YES
NO
Input
8 3 3
1 2
2 3
3 4
4 5
4 6
6 7
7 8
2 5 8
2
7 1
8 1
Output
YES
NO
Note
The graph for the first example is shown below. The rest stop is denoted by red.
For the first query, Bessie can visit these cities in order: 1, 2, 3.
For the second query, Bessie can visit these cities in order: 3, 2, 4, 5.
For the third query, Bessie cannot travel to her destination. For example, if she attempts to travel this way: 3, 2, 4, 5, 6, she travels on more than 2 roads without resting.
<image>
The graph for the second example is shown below.
<image>
Submitted Solution:
```
import sys,io
sys.setrecursionlimit(10**6)
def Check(y,rest,rcities,arr,k,y2):
if(y in rcities):
rcities.remove(y)
rest.append(y)
return
if(k==0):
#print("k is 0")
return
else:
p=int()
arr2=arr[y-1].copy()
if(len(arr2)>1):
try:
index=int(arr2.index(y2))
arr2=arr2[:index]+arr2[index+1:]
except:
pass
for i in arr2:
Check(i,rest,rcities,arr,k-1,y)
if(len(rcities)==0):
break
def DFS(x,y,rcities,arr,k,k2,x2):
# print('x= ',x,"y= ",y,"k= ",k,"\narr in func=\n")
#for row in arr:
# print(*row)
# print("rcities= ",*rcities)
if(x==y):
return -1
if(x in rcities):
rcities.remove(x)
k=k2
if(k==0):
#print("k is 0")
return 0
else:
p=int()
arr2=arr[x-1].copy()
if(len(arr2)>1):
try:
index=int(arr2.index(x2))
arr2=arr2[:index]+arr2[index+1:]
except:
pass
for i in arr2:
p=DFS(i,y,rcities,arr,k-1,k2,x)
if(p==-1):
return -1
if(p!=0):
return p
if(p==0):
# print("YO")
return 0
n,k,r = map(int,input().split())
arr = [[] for i in range (n)]
for i in range(n-1):
x,y =map(int,input().split())
arr[x-1].append(y)
arr[y-1].append(x)
#print("arr in main=")
#for row in arr:
# print(*row)
rcities=[]
rcities=list(map(int,input().split()))
v=int(input())
for i in range(v):
x,y = map(int,input().split())
rcities2=rcities.copy()
rcities3=rcities.copy()
rest = list()
Check(y,rest,rcities3,arr,k,y)
if(len(rest)==0):
buffer = io.BytesIO()
sys.stdout.buffer.write(b"NO\n")
del buffer
continue
p=DFS(x,y,rcities2,arr,k,k,x)
if(p==-1):
buffer = io.BytesIO()
sys.stdout.buffer.write(b"YES\n")
del buffer
else:
buffer = io.BytesIO()
sys.stdout.buffer.write(b"NO\n")
del buffer
``` | instruction | 0 | 56,522 | 1 | 113,044 |
No | output | 1 | 56,522 | 1 | 113,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bessie is planning a vacation! In Cow-lifornia, there are n cities, with n-1 bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city.
Bessie is considering v possible vacation plans, with the i-th one consisting of a start city a_i and destination city b_i.
It is known that only r of the cities have rest stops. Bessie gets tired easily, and cannot travel across more than k consecutive roads without resting. In fact, she is so desperate to rest that she may travel through the same city multiple times in order to do so.
For each of the vacation plans, does there exist a way for Bessie to travel from the starting city to the destination city?
Input
The first line contains three integers n, k, and r (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k,r ≤ n) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.
Each of the following n-1 lines contain two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), meaning city x_i and city y_i are connected by a road.
The next line contains r integers separated by spaces — the cities with rest stops. Each city will appear at most once.
The next line contains v (1 ≤ v ≤ 2 ⋅ 10^5) — the number of vacation plans.
Each of the following v lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the start and end city of the vacation plan.
Output
If Bessie can reach her destination without traveling across more than k roads without resting for the i-th vacation plan, print YES. Otherwise, print NO.
Examples
Input
6 2 1
1 2
2 3
2 4
4 5
5 6
2
3
1 3
3 5
3 6
Output
YES
YES
NO
Input
8 3 3
1 2
2 3
3 4
4 5
4 6
6 7
7 8
2 5 8
2
7 1
8 1
Output
YES
NO
Note
The graph for the first example is shown below. The rest stop is denoted by red.
For the first query, Bessie can visit these cities in order: 1, 2, 3.
For the second query, Bessie can visit these cities in order: 3, 2, 4, 5.
For the third query, Bessie cannot travel to her destination. For example, if she attempts to travel this way: 3, 2, 4, 5, 6, she travels on more than 2 roads without resting.
<image>
The graph for the second example is shown below.
<image>
Submitted Solution:
```
def DFS(x,y,rcities,arr,k):
# print('x= ',x,"y= ",y,"k= ",k,"\narr in func=\n")
#for row in arr:
# print(*row)
if(x==y):
return -1
if(x in rcities):
rcities.remove(x)
return x
if(k==0):
# print("k is 0")
return 0
else:
p=int()
for i in range(n):
if(arr[x-1][i]==1):
p=DFS(i+1,y,rcities,arr,k-1)
if(p==-1):
return -1
if(p!=0):
return p
if(p==0):
#print("YO")
return 0
def Check(x,y,rcities,arr,k):
p=int()
while(1):
p = DFS(x,y,rcities,arr,k)
if(p==0):
return "NO"
if(p==-1):
return "YES"
else:
x = p
n,k,r = map(int,input().split())
arr=[[0 for j in range(n)]for i in range(n)]
for i in range(n-1):
x,y =map(int,input().split())
arr[x-1][y-1]=1
arr[y-1][x-1]=1
#print("arr in main=")
#for row in arr:
# print(*row)
rcities=[]
rcities=list(map(int,input().split()))
v=int(input())
for i in range(v):
x,y = map(int,input().split())
rcities2=rcities.copy()
p=Check(x,y,rcities2,arr,k)
print(p)
``` | instruction | 0 | 56,523 | 1 | 113,046 |
No | output | 1 | 56,523 | 1 | 113,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,645 | 1 | 113,290 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
n, m = map(int, input().split())
road = [[] for i in range(m)]
for i in range(m):
road[i] = [i] + list(map(int, input().split())) + ['NONE']
for i in road:
if i[2] < i[1]:
i[1], i[2] = i[2], i[1]
i[1], i[2] = i[1] - 1, i[2] - 1
participation = [[] for i in range(m)]
for i in range(len(road)):
for j in range(i + 1, len(road)):
if (road[j][1] < road[i][1] < road[j][2]) ^ (road[j][1] < road[i][2] < road[j][2]):
if road[j][1] != road[i][1] and road[j][2] != road[i][1] and road[j][1] != road[i][2] and road[j][2] != road[i][2]:
participation[i].append(j)
participation[j].append(i)
result = ""
mark = [0] * m
stack = []
while sum(mark) != m:
if len(stack) == 0:
for i in range(len(mark)):
if mark[i] == 0:
stack.append(i)
break
index = stack.pop()
mark[index] = 1
if road[index][3] == "NONE":
road[index][3] = "i"
for i in participation[index]:
if road[i][3] == road[index][3]:
result = "Impossible"
print(result)
break
elif road[index][3] != "i" and road[i][3] == "NONE":
road[i][3] = "i"
stack.append(i)
elif road[index][3] == "i" and road[i][3] == "NONE":
road[i][3] = "o"
stack.append(i)
if result == "Impossible":
break
if result != "Impossible":
for i in road:
result += i[3]
print(result)
``` | output | 1 | 56,645 | 1 | 113,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,646 | 1 | 113,292 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin, stdout, stderr
from collections import deque
from time import clock
def intersect(x,y):
if x[0] < y[0] and y[0] < x[1] and x[1] < y[1]:
return True
elif y[0] < x[0] and x[0] < y[1] and y[1] < x[1]:
return True
else:
return False
n,m = map(int,stdin.readline().split())
adj = [[] for i in range(2*m)]
roads = []
for i in range(m):
a,b = map(int,stdin.readline().split())
roads.append((min(a,b),max(a,b)))
for j in range(i):
if intersect(roads[i], roads[j]):
adj[2*i].append(2*j+1)
adj[2*j].append(2*i+1)
adj[2*i+1].append(2*j)
adj[2*j+1].append(2*i)
stderr.write('input: {}\n'.format(clock()))
idx = [None for i in range(2*m)]
low = [None for i in range(2*m)]
num = [None for i in range(2*m)]
stk = deque()
cnum = 0
def scc(cur):
global idx, low, num, stk, cnum
if idx[cur] is not None:
return low[cur]
idx[cur] = low[cur] = cur
stk.append(cur)
for x in adj[cur]:
low[cur] = min(low[cur], scc(x))
if idx[cur] == low[cur]:
add = stk.pop()
num[add] = cnum
while add != cur:
add = stk.pop()
num[add] = cnum
cnum += 1
return low[cur]
for i in range(2*m):
if idx[i] is None:
scc(i)
stderr.write('scc: {}\n'.format(clock()))
ans = []
for i in range(m):
if num[2*i] == num[2*i+1]:
stdout.write('Impossible\n')
break
else:
if num[2*i] < num[2*i+1]:
ans.append('i')
else:
ans.append('o')
else:
stdout.write(''.join(ans)+'\n')
``` | output | 1 | 56,646 | 1 | 113,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,647 | 1 | 113,294 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
#!/usr/bin/env python3
# Read inputs
(n,m) = map(int, input().split())
roads = []
for i in range(m):
(a,b) = map(int, input().split())
a,b = a-1,b-1
roads.append((min(a,b),max(a,b)))
# Check for incompatibilities between roads
incompatible = [[] for i in range(m)]
for i in range(m):
(a,b) = roads[i]
side1 = set(range(a+1,b))
side2 = set(range(n)).difference(side1)
side2.remove(a)
side2.remove(b)
for j in range(i+1,m):
# Check if the roads are compatible
(a,b) = roads[j]
problem = ((a in side1) and (b in side2)) or ((a in side2) and (b in side1))
if (problem):
incompatible[i].append(j)
incompatible[j].append(i)
# Decide where the roads should go
decision = [None for i in range(m)]
def update(decision, road_idx, value):
if (decision[road_idx] == value):
# Nothing to do
pass
elif (decision[road_idx] == (not value)):
# Problem
print("Impossible")
exit()
else:
decision[road_idx] = value;
for other_road in incompatible[road_idx]:
update(decision, other_road, not value)
for i in range(m):
if (decision[i] == None):
# Just pick one
update(decision, i, True)
output = ["i" if d else "o" for d in decision]
print("".join(output))
``` | output | 1 | 56,647 | 1 | 113,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,648 | 1 | 113,296 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, m = map(int, input().split())
edges = [sorted(map(lambda x: int(x) - 1, input().split())) for _ in range(m)]
adj = [[] for _ in range(m)]
for i, (u1, v1) in enumerate(edges):
for j, (u2, v2) in enumerate(edges[i + 1:], start=i + 1):
if u1 in (u2, v2) or v1 in (u2, v2):
continue
if (1 if u1 < u2 < v1 else 0) ^ (1 if u1 < v2 < v1 else 0) == 1:
adj[i].append(j)
adj[j].append(i)
color = [-1] * m
for i in range(m):
if color[i] != -1:
continue
color[i] = 0
stack = [i]
while stack:
v = stack.pop()
for dest in adj[v]:
if color[dest] != -1 and color[dest] == color[v]:
print('Impossible')
exit()
if color[dest] == -1:
color[dest] = color[v] ^ 1
stack.append(dest)
ans = ''.join('o' if color[i] else 'i' for i in range(m))
print(ans)
``` | output | 1 | 56,648 | 1 | 113,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,649 | 1 | 113,298 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
from collections import defaultdict
import sys
class Graph:
def __init__(self,vertices):
self.V= vertices # Nro de Vertices
self.graph = defaultdict(list) # Diccionario de adyacentes
def getVertices(self):
return self.V
# Agregar una arista
def addEdge(self,u,v):
self.graph[u].append(v)
# Funcion que dispara el dfs y asigna el numero de SCC
def DFSUtil(self,v,visited, comp, comp_act):
# Marca el nodo actual como visitado
visited[v]= True
# Asigna el numero de SCC actual a el nodo actual
comp[v]=comp_act
# Recurrencia sobre los nodos adyacentes
for i in self.graph[v]:
if visited[i]==False:
self.DFSUtil(i,visited, comp, comp_act)
def fillOrder(self,v,visited, stack):
# Mark the current node as visited
visited[v]= True
#Recur for all the vertices adjacent to this vertex
for i in self.graph[v]:
if visited[i]==False:
self.fillOrder(i, visited, stack)
stack = stack.append(v)
# Funcion que retorna el grafo
def getTranspose(self):
g = Graph(self.V)
for i in self.graph:
for j in self.graph[i]:
g.addEdge(j,i)
return g
# Funcion principal para encontrar las SCCs
def getSCCs(self):
stack = []
# Marcar todos los vertices como no visitados (para el primer dfs)
visited =[False]*(self.V)
# Carga la pila con los vertices usando un dfs
for i in range(self.V):
if visited[i]==False:
self.fillOrder(i, visited, stack)
# Crea el grafo invertido
gr = self.getTranspose()
# Marcar todos los vertices como no visitados (para el segundo dfs)
visited =[False]*(self.V)
# Vector que indicara la SCC de cada vertice
comp = [-1]*(self.V)
comp_act = 0
# Finalmente procesa todos los vertices (sobre el grafo invertido) en el orden que se insertaron en la pila
while stack:
i = stack.pop()
if visited[i]==False:
gr.DFSUtil(i, visited, comp, comp_act)
comp_act+=1
return comp, comp_act
def getNot(cantRoads, road):
if road > cantRoads -1:
return road - cantRoads
else:
return road + cantRoads
def addClause(graph, cantR, a, b):
graph.addEdge(getNot(cantR, a), b)
graph.addEdge(getNot(cantR, b), a)
# Programa principal
lis = list(map(lambda x:int(x), input().split(' ')))
cantCities = lis[0]
cantRoads = lis[1]
graph = Graph(2*cantRoads)
# Lee las rutas
r = []
for i in range(cantRoads):
road = list(map(lambda x:int(x), input().split(' ')))
# Mantener el orden para no evitar entrecruces despues
if road[0] > road[1]:
road[1], road[0] = road[0], road[1]
r.append( (road[0] - 1, road[1] - 1) )
for i in range(cantRoads):
for j in range(cantRoads):
if (i != j):
# Si no se interceptan
if (r[i][0] < r[j][0]) and (r[i][1] > r[j][0]) and (r[i][1] < r[j][1]):
# Tengo que añadir (i o j) y (~i o ~j) para decir que no pueden estar en el mismo valor
# La funcion addClause() agrega la clausula equivalente al grafo
addClause(graph, cantRoads, i, j)
addClause(graph, cantRoads, getNot(cantRoads, i), getNot(cantRoads, j))
# Ahora ya podemos encontrar las componentes fuertemente conexas
comp, cantComp = graph.getSCCs()
# Verificar solucion
for i in range(cantRoads):
# Si una ruta y su negacion estan en la misma SCC entonces no hay solucion
if(comp[i] == comp[getNot(cantRoads, i)]):
print("Impossible")
sys.exit()
# Ahora hay que asignarle un valor a las rutas (dentro o fuera)
res = [-1]*cantComp
for i in range(graph.getVertices()): #Por cada SCC
if (res[comp[i]]) == -1: # Si la i no tiene valor
res[comp[i]] = True # Le asigna true
res[comp[getNot(cantRoads, i)]] = False
s = ""
for i in range(cantRoads):
if(res[comp[i]]):
s = s + "o"
else:
s = s + "i"
print(s)
``` | output | 1 | 56,649 | 1 | 113,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,650 | 1 | 113,300 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
def intersect(mnn, mxx, crx, cry):
if mnn <= crx and mxx >= cry or crx <= mnn and cry >= mxx:
return False
if cry <= mxx and mnn <= crx and cry <= crx:
return False
if mnn >= cry and mnn >= crx or mxx <= crx and mxx <= cry:
return False
return True
def dfs(visit,graph,cr,vis):
#print(visit)
if visit[cr] != -1 and visit[cr] != vis:
print('Impossible')
exit()
if visit[cr]!=-1:
return
visit[cr] = vis
for u in graph[cr]:
dfs(visit,graph,u, 1 - vis)
def main():
n,m = map(int,input().strip().split())
inn=[]
visit = [-1]*m
graph = [[] for _ in range(m)]
for i in range(m):
x,y = map(int,input().strip().split())
mn,mx = min(x,y), max(x,y)
for j,edge in enumerate(inn):
if intersect(mn,mx,edge[0],edge[1]):
graph[i].append(j)
graph[j].append(i)
inn.append((mn,mx))
#print(graph)
#print(visit)
for i in range(m):
if visit[i] == -1:
dfs(visit,graph,i,0)
print("".join(['o' if i == 0 else 'i' for i in visit]))
if __name__ == "__main__":
main()
"""
6 3
1 4
2 5
3 6
5 3
1 3
2 5
1 4
4 2
1 3
2 4
6 3
1 3
3 5
5 1
"""
``` | output | 1 | 56,650 | 1 | 113,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,651 | 1 | 113,302 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
def intersect(l1, r1, l2, r2):
if l1 <= l2 and r1 >= r2 or l2 <= l1 and r2 >= r1:
return False
if r2 <= r1 and l1 <= l2 and r2 <= l2:
return False
if l1 >= r2 and l1 >= l2 or r1 <= l2 and r1 <= r2:
return False
return True
def ans():
print(''.join(['i' if i == 0 else 'o' for i in color]))
def dfs(v, clr):
if color[v] != -1 and color[v] != clr:
print('Impossible')
exit(0)
if color[v] != -1:
return
color[v] = clr
for u in graph[v]:
dfs(u, 1 - clr)
n, m = map(int, input().split())
graph = [[] for i in range(m)]
vedge = []
color = [-1] * m
for i in range(m):
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
for j, v in enumerate(vedge):
if intersect(a, b, v[0], v[1]):
graph[i].append(j)
graph[j].append(i)
vedge.append((a, b))
for i in range(m):
if color[i] == -1:
dfs(i, 0)
ans()
``` | output | 1 | 56,651 | 1 | 113,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo | instruction | 0 | 56,652 | 1 | 113,304 |
Tags: 2-sat, dfs and similar, dsu, graphs
Correct Solution:
```
from collections import defaultdict
import sys
class Graph:
def __init__(self,vertices):
self.V= vertices # Nro de Vertices
self.graph = defaultdict(list) # Diccionario
def getVertices(self):
return self.V
# Agregar una arista
def addEdge(self,u,v):
self.graph[u].append(v)
# Funcion que dispara el dfs y asigna el numero de SCC
def DFSUtil(self,v,visited, comp, comp_act):
# Marca el nodo actual como visitado
visited[v]= True
# Asigna el numero de SCC actual a el nodo actual
comp[v]=comp_act
# Recurrencia sobre los nodos adyacentes
for i in self.graph[v]:
if visited[i]==False:
self.DFSUtil(i,visited, comp, comp_act)
def fillOrder(self,v,visited, stack):
# Mark the current node as visited
visited[v]= True
#Recur for all the vertices adjacent to this vertex
for i in self.graph[v]:
if visited[i]==False:
self.fillOrder(i, visited, stack)
stack = stack.append(v)
# Funcion que retorna el grafo
def getTranspose(self):
g = Graph(self.V)
for i in self.graph:
for j in self.graph[i]:
g.addEdge(j,i)
return g
# Funcion principal para encontrar las SCCs
def getSCCs(self):
stack = []
# Marcar todos los vertices como no visitados (para el primer dfs)
visited =[False]*(self.V)
# Carga la pila con los vertices usando un dfs
for i in range(self.V):
if visited[i]==False:
self.fillOrder(i, visited, stack)
# Crea el grafo invertido
gr = self.getTranspose()
# Marcar todos los vertices como no visitados (para el segundo dfs)
visited =[False]*(self.V)
# Vector que indicara la SCC de cada vertice
comp = [-1]*(self.V)
comp_act = 0
# Finalmente procesa todos los vertices (sobre el grafo invertido) en el orden que se insertaron en la pila
while stack:
i = stack.pop()
if visited[i]==False:
gr.DFSUtil(i, visited, comp, comp_act)
comp_act+=1
return comp, comp_act
def getNot(cantRoads, road):
if road > cantRoads -1:
return road - cantRoads
else:
return road + cantRoads
def addClause(graph, cantR, a, b):
graph.addEdge(getNot(cantR, a), b)
graph.addEdge(getNot(cantR, b), a)
# Programa principal
lis = list(map(lambda x:int(x), input().split(' ')))
cantCities = lis[0]
cantRoads = lis[1]
graph = Graph(2*cantRoads)
# Lee las rutas
r = []
for i in range(cantRoads):
road = list(map(lambda x:int(x), input().split(' ')))
# Mantener el orden para no evitar entrecruces despues
if road[0] > road[1]:
road[1], road[0] = road[0], road[1]
r.append( (road[0] - 1, road[1] - 1) )
for i in range(cantRoads):
for j in range(cantRoads):
if (i != j):
# Si no se interceptan
if (r[i][0] < r[j][0]) and (r[i][1] > r[j][0]) and (r[i][1] < r[j][1]):
# Tengo que añadir (i o j) y (~i o ~j) para decir que no pueden estar en el mismo valor
# La funcion addClause() agrega la clausula equivalente al grafo
addClause(graph, cantRoads, i, j)
addClause(graph, cantRoads, getNot(cantRoads, i), getNot(cantRoads, j))
# Ahora ya podemos encontrar las componentes fuertemente conexas
comp, cantComp = graph.getSCCs()
for i in range(cantRoads):
# Si una ruta y su negacion estan en la misma SCC entonces no hay solucion
if(comp[i] == comp[getNot(cantRoads, i)]):
print("Impossible")
sys.exit()
# Ahora hay que asignarle un valor a las rutas (dentro o fuera)
res = [-1]*graph.getVertices()
for i in range(cantComp):
if (res[i]) == -1:
res[comp[i]] = True
a = getNot(cantRoads, i)
res[comp[a]] = False
s = ""
for i in range(cantRoads):
if(res[comp[i]]):
s = s + "o"
else:
s = s + "i"
print(s)
``` | output | 1 | 56,652 | 1 | 113,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo
Submitted Solution:
```
n, m = list(map(int, input().split()))
inS = []
outS = []
string = ""
for _ in range(m):
s, f = list(map(int, input().split()))
if s > f:
f, s = s, f
flag = False
for i in inS:
if (i[0] <= s <= f <= i[1]) or (s < i[0] and f > i[1]) or (s == i[0] or s == i[1]) or (f == i[1] or f == i[0]):
pass
else:
flag = True
break
if flag:
flag = False
for i in outS:
if (i[0] <= s <= f <= i[1]) or (s < i[0] and f > i[1]) or (s == i[0] or s == i[1]) or (
f == i[1] or f == i[0]):
pass
else:
flag = True
break
if not flag:
outS.append((s,f))
string += "o"
else:
print("Impossible")
exit()
else:
inS.append((s,f))
string += "i"
print(string)
``` | instruction | 0 | 56,653 | 1 | 113,306 |
No | output | 1 | 56,653 | 1 | 113,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo
Submitted Solution:
```
n,m = map(int,input().split(' '))
ans = []
for i in range(0,m):
a,b = map(int,input().split(' '))
if a%2==0 and b%2==0:
ans.append('i')
else:
ans.append('o')
ans = ''.join(ans)
print(ans)
``` | instruction | 0 | 56,654 | 1 | 113,308 |
No | output | 1 | 56,654 | 1 | 113,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).
Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?
Input
The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.
Output
If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.
Examples
Input
4 2
1 3
2 4
Output
io
Input
6 3
1 3
3 5
5 1
Output
ooo
Submitted Solution:
```
n, m = list(map(int, input().split()))
inS = []
outS = []
string = ""
for _ in range(m):
s, f = list(map(int, input().split()))
if s > f:
f, s = s, f
flag = False
for i in inS:
if (i[0] <= s <= f <= i[1]) or (s < i[0] and f > i[1]) or (s <= f <= i[0]) or (i[1] <= s <= f) or (s == i[0] or s == i[1]) or (f == i[1] or f == i[0]):
pass
else:
flag = True
break
if flag:
flag = False
for i in outS:
if (i[0] <= s <= f <= i[1]) or (s < i[0] and f > i[1]) or (s == i[0] or s == i[1]) or (
f == i[1] or f == i[0]):
pass
else:
flag = True
break
if not flag:
outS.append((s,f))
string += "o"
else:
print("Impossible")
exit()
else:
inS.append((s,f))
string += "i"
print(string)
``` | instruction | 0 | 56,655 | 1 | 113,310 |
No | output | 1 | 56,655 | 1 | 113,311 |
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