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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
#!/usr/bin/python3
import sys
from fractions import Fraction
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().strip().split()]
a.sort()
s1 = sum(a)
p = [0 for i in range(n)]
c = [1 for i in range(n)]
for i in range(1, n):
p[i] = (a[i] - a[i-1]) * c[i-1] + p[i-1]
c[i] += c[i-1]
s2 = 2*sum(p)
ans = Fraction(s1 + s2, n)
print("{0} {1}".format(ans.numerator, ans.denominator))
``` | instruction | 0 | 92,999 | 1 | 185,998 |
Yes | output | 1 | 92,999 | 1 | 185,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
from math import gcd
n = int(input())
a = sorted(map(int, input().split()))
ans = 0
for i in range(n):
ans += 2 * (i + i - n + 1) * a[i]
ans += sum(a)
tmp = gcd(ans, n)
print(ans // tmp, n // tmp)
``` | instruction | 0 | 93,000 | 1 | 186,000 |
Yes | output | 1 | 93,000 | 1 | 186,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 3 11:57:46 2020
@author: shailesh
"""
from math import gcd
def reduce_fraction(x,y):
d = gcd(x,y)
x = x//d
y = y//d
return x,y
N = int(input())
A = [int(i) for i in input().split()]
A.sort()
d0 = A[0]
sum_val = 0
for i in range(N-1):
m_bf = i+2
m_af = N - i - 1
d = A[i+1]-A[i]
# d = 1
sum_val +=m_af*(2*m_bf - 1)*d
# print(A[i],A[i+1],sum_val)
numerator = N*d0 + sum_val
denominator = N
numerator,denominator = reduce_fraction(numerator,denominator)
print(numerator,denominator)
#from itertools import permutations
#perms = list(permutations([2,3,5]))
#
#perms = [(0,) + perm for perm in perms]
#
#d = {}
#d['02'] = 0
#d['23'] = 0
#d['35'] = 0
#for perm in perms:
# for i in range(len(perm)-1):
#
# start_end = [perm[i],perm[i+1]]
# start_end.sort()
# rng = range(start_end[0],start_end[1]+1)
# if 0 in rng and 2 in rng:
# d['02'] +=1
# if 2 in rng and 3 in rng:
# d['23'] += 1
# if 3 in rng and 5 in rng:
# d['35'] +=1
``` | instruction | 0 | 93,001 | 1 | 186,002 |
Yes | output | 1 | 93,001 | 1 | 186,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
from fractions import Fraction
while(1):
try:
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
a.sort()
for i in range(0,n):
s+=2*(2*i+1-n)*a[i]
f=Fraction(s,n)
print(f.numerator,end=' ')
print(f.denominator)
except EOFError:
break
``` | instruction | 0 | 93,002 | 1 | 186,004 |
Yes | output | 1 | 93,002 | 1 | 186,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from itertools import permutations
from math import gcd
def main():
n = int(input())
a = list(map(int, input().split()))
cnt = 0
asum = 0
for i in permutations(a):
print(i)
asum += i[0]
cnt += 1
for j in range(1, n):
asum += abs(i[j-1] - i[j])
print(asum // gcd(asum, cnt), cnt // gcd(asum, cnt))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,003 | 1 | 186,006 |
No | output | 1 | 93,003 | 1 | 186,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
from fractions import Fraction
n = int(input())
l = [int(x) for x in input().split()]
f = Fraction(sum([(3-2*n+4*i)*l[i] for i in range(n)]),n)
print(f.numerator,f.denominator)
``` | instruction | 0 | 93,004 | 1 | 186,008 |
No | output | 1 | 93,004 | 1 | 186,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
from itertools import accumulate
# from collections import defaultdict, Counter
import math
def modinv(n,p):
return pow(n,p-2,p)
def main():
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
n = int(input())
arr = [int(x) for x in input().split()]
den = 2 ** n - 2
num = 0
s = sum(arr)
pre_sum = list(accumulate(arr))
for i in range(n):
left = 0
right = 0
if i > 0:
left = pre_sum[i-1]
if i < n-1:
right = s - pre_sum[i]
if right > 0:
right -= (arr[i] * (n-i-1))
if left > 0:
left = (i * arr[i]) - left
num += (left + arr[i] + right)*(n-1)
# print(c)
g = math.gcd(num, den)
print(num//g, den//g)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | instruction | 0 | 93,005 | 1 | 186,010 |
No | output | 1 | 93,005 | 1 | 186,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 โค n โค 105). Next line contains n distinct integers a1, a2, ..., an (1 โค ai โค 107).
Output
Output two integers โ the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
* [2, 3, 5]: total distance traveled: |2 โ 0| + |3 โ 2| + |5 โ 3| = 5;
* [2, 5, 3]: |2 โ 0| + |5 โ 2| + |3 โ 5| = 7;
* [3, 2, 5]: |3 โ 0| + |2 โ 3| + |5 โ 2| = 7;
* [3, 5, 2]: |3 โ 0| + |5 โ 3| + |2 โ 5| = 8;
* [5, 2, 3]: |5 โ 0| + |2 โ 5| + |3 โ 2| = 9;
* [5, 3, 2]: |5 โ 0| + |3 โ 5| + |2 โ 3| = 8.
The average travel distance is <image> = <image> = <image>.
Submitted Solution:
```
import math
n = int(input())
l = [int(x) for x in input().split()]
a1 = sum(l)
a2 = n
a3 = 0
temp = 0
for i in range(n):
temp += l[n-i-1]
a3-=(a1-temp)
a3+=(n-i-1)*(l[n-i-1])
a1 = a1+a3+a3
a4 = math.gcd(a1, a2)
print(a1//a4, a2//a4)
``` | instruction | 0 | 93,006 | 1 | 186,012 |
No | output | 1 | 93,006 | 1 | 186,013 |
Provide tags and a correct Python 2 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,235 | 1 | 186,470 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
n,m,s=in_arr()
d=[[] for i in range(n+1)]
for i in range(m):
u,v=in_arr()
d[u].append(v)
good=[False]*(n+1)
q=[s]
good[s]=True
while q:
x=q.pop()
for i in d[x]:
if not good[i]:
good[i]=True
q.append(i)
arr=[]
for i in range(1,n+1):
if not good[i]:
vis=list(good)
q=[i]
c=1
vis[i]=1
while q:
x=q.pop()
for j in d[x]:
if not vis[j]:
vis[j]=True
q.append(j)
c+=1
arr.append((c,i))
arr.sort(reverse=True)
ans=0
for _,i in arr:
if not good[i]:
good[i]=True
q=[i]
ans+=1
while q:
x=q.pop()
for j in d[x]:
if not good[j]:
good[j]=1
q.append(j)
pr_num(ans)
``` | output | 1 | 93,235 | 1 | 186,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,236 | 1 | 186,472 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
from collections import defaultdict
import sys
sys.setrecursionlimit(1000000)
n, m, s = map(int, input().split())
s = s - 1
def read_graph():
g = defaultdict(list)
for _ in range(m):
(u, v) = map(lambda x: int(x) - 1, input().split())
if v != s:
g[u].append(v)
return g
G = read_graph()
vis = defaultdict(lambda: False)
topo = []
def dfs(u):
# print(u)
for v in G[u]:
if not vis[v]:
vis[v] = True
dfs(v)
topo.append(u)
for i in range(n):
if not vis[i]:
vis[i] = True
dfs(i)
vis.clear()
vis[s] = True
dfs(s)
ans = 0
for i in topo[::-1]:
if not vis[i]:
vis[i] = True
ans += 1
dfs(i)
print(ans)
``` | output | 1 | 93,236 | 1 | 186,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,237 | 1 | 186,474 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
import sys
n, m, s = map(int, input().split())
adj = [[] for _ in range(500005)]
ar = []
vis = [0] * 500005
sys.setrecursionlimit(6000)
def dfs(s):
vis[s] = 1
for i in range(len(adj[s])):
if(vis[adj[s][i]] == 0):
dfs(adj[s][i])
ar.append(s)
for i in range(m):
u, v = map(int, input().split())
adj[u].append(v)
dfs(s)
for i in range(n):
if(vis[i + 1] == 0):
dfs(i + 1)
res = 0
vis = [0] * 500005
for i in range(n - 1, -1, -1):
if(vis[ar[i]] == 0):
if(s != ar[i]):
res += 1
dfs(ar[i])
print(res)
``` | output | 1 | 93,237 | 1 | 186,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,238 | 1 | 186,476 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
from __future__ import print_function
from queue import Queue
from random import shuffle
import sys
import math
import os.path
sys.setrecursionlimit(100000)
# LOG
def log(*args, **kwargs):
print(*args, file=sys.stderr, **kwargs)
# INPUT
def ni():
return map(int, input().split())
def nio(offset):
return map(lambda x: int(x) + offset, input().split())
def nia():
return list(map(int, input().split()))
# CONVERT
def toString(aList, sep=" "):
return sep.join(str(x) for x in aList)
def toMapInvertIndex(aList):
return {k: v for v, k in enumerate(aList)}
# SORT
def sortId(arr):
return sorted(range(len(arr)), key=lambda k: arr[k])
# MAIN
n,m,s = ni()
s -= 1
adj = [[] for _ in range(n)]
for i in range(m):
u,v = nio(-1)
if (v != s):
adj[u].append(v)
stack = []
visited= [False]*n
def dfs(x):
global visited
global stack
visited[x] = True
for y in adj[x]:
if not visited[y]:
dfs(y)
stack.append(x)
for i in range(n):
if not visited[i]:
dfs(i)
# log(adj)
# log(visited)
# log(stack)
count = -1
def loang(x):
global visited
visited[x] = False
for y in adj[x]:
if visited[y]:
loang(y)
for x in stack[::-1]:
if visited[x]:
count += 1
loang(x)
print(count)
``` | output | 1 | 93,238 | 1 | 186,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,239 | 1 | 186,478 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
import sys
sys.setrecursionlimit(10**4)
from collections import deque
n, m, s = map(int, input().split())
s -= 1
graph = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
graph[u-1].append(v-1)
seen = [False]*n
li = deque()
def visit(node, t=True):
if not seen[node]:
seen[node] = True
for c_node in graph[node]:
visit(c_node)
if t:
li.appendleft(node)
for i in range(n):
visit(i)
seen = [False]*n
cnt = 0
visit(s, False)
for i in list(li):
if seen[i]:
continue
visit(i, False)
cnt += 1
print(cnt)
``` | output | 1 | 93,239 | 1 | 186,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,240 | 1 | 186,480 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
from collections import defaultdict
import sys
def dfs(u):
avail[u] = False
for v in g[u]:
if avail[v]:
dfs(v)
topo.append(u)
sys.setrecursionlimit(6000)
n, m, s = map(int, input().split())
g = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
g[u - 1].append(v - 1)
avail, topo = [True] * n, []
for i,a in enumerate(avail):
if a:
dfs(i)
avail, res = [True] * n, 0
dfs(s - 1)
for i in reversed(topo):
if avail[i]:
res += 1
dfs(i)
print(res)
``` | output | 1 | 93,240 | 1 | 186,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,241 | 1 | 186,482 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
def main():
import sys
sys.setrecursionlimit(10**5)
from collections import deque
n, m, s = map(int, input().split())
s -= 1
graph = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
graph[u-1].append(v-1)
seen = [False]*n
li = deque()
def visit(node):
if not seen[node]:
seen[node] = True
for c_node in graph[node]:
visit(c_node)
li.appendleft(node)
def visit2(node):
if not seen[node]:
seen[node] = True
for c_node in graph[node]:
visit2(c_node)
for i in range(n):
visit(i)
seen = [False]*n
cnt = 0
visit2(s)
for i in li:
if seen[i]:
continue
visit2(i)
cnt += 1
print(cnt)
if __name__ == "__main__":
try:
main()
except:
print('error!')
exit(0)
``` | output | 1 | 93,241 | 1 | 186,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,242 | 1 | 186,484 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n, m, s = map(int, input().split())
s -= 1
adj = [[] for _ in range(n)]
rev = [[] for _ in range(n)]
for u, v in (map(int, input().split()) for _ in range(m)):
adj[u-1].append(v-1)
rev[v-1].append(u-1)
group = [-1]*n
g = 0
group[s] = g
stack = [s]
while stack:
v = stack.pop()
for dest in adj[v]:
if group[dest] != -1:
continue
group[dest] = g
stack.append(dest)
for i in range(n):
if group[i] != -1:
continue
g += 1
group[i] = g
stack = [i]
while stack:
v = stack.pop()
for dest in adj[v]:
if group[dest] == 0 or group[dest] == g:
continue
group[dest] = g
stack.append(dest)
print(len(set(group)) - 1)
``` | output | 1 | 93,242 | 1 | 186,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5. | instruction | 0 | 93,243 | 1 | 186,486 |
Tags: dfs and similar, graphs, greedy
Correct Solution:
```
import sys
sys.setrecursionlimit(100001)
from collections import *
d=deque()
def dfs2(start):
vis[start]=1
for child in adj[start]:
if vis[child]==0:
dfs2(child)
def dfs(start):
vis[start]=1
for child in adj[start]:
if vis[child]==0:
dfs(child)
d.appendleft(start)
n,m,k=map(int,input().split())
adj=[]
for i in range(n):
adj.append([])
for i in range(m):
a,b=map(int,input().split())
adj[a-1].append(b-1)
vis=[0]*n
for i in range(n):
if vis[i]==0:
dfs(i)
vis=[0]*n
dfs2(k-1)
cnt=0
for i in d:
if vis[i]==0:
dfs2(i)
cnt+=1
print(cnt)
``` | output | 1 | 93,243 | 1 | 186,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
import sys
def d(u):
rt[u] = False
for v in q[u]:
if rt[v]:
d(v)
topo.append(u)
sys.setrecursionlimit(6000)
n, m, s = map(int, input().split())
q = [[] for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
q[u - 1].append(v - 1)
rt, topo = [True] * n, []
for i,a in enumerate(rt):
if a:
d(i)
rt, res = [True] * n, 0
d(s - 1)
for i in reversed(topo):
if rt[i]:
res += 1
d(i)
print(res)
``` | instruction | 0 | 93,244 | 1 | 186,488 |
Yes | output | 1 | 93,244 | 1 | 186,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
import sys
n, m, s = map(int, input().split())
adj = [[] for i in range(500005)]
ar = []
vis = [0 for i in range(500005)]
sys.setrecursionlimit(6000)
def dfs(s):
vis[s] = 1
for i in range(len(adj[s])):
if(vis[adj[s][i]] == 0):
dfs(adj[s][i])
ar.append(s)
for i in range(m):
u, v = map(int, input().split())
adj[u].append(v)
dfs(s)
for i in range(n):
if(vis[i + 1] == 0):
dfs(i + 1)
res = 0
vis = [0 for i in range(500005)]
for i in range(n - 1, -1, -1):
if(vis[ar[i]] == 0):
if(s != ar[i]):
res += 1
dfs(ar[i])
print(res)
``` | instruction | 0 | 93,245 | 1 | 186,490 |
Yes | output | 1 | 93,245 | 1 | 186,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
from collections import deque
import heapq
import sys
input = sys.stdin.readline
def bfs1(s):
q = deque()
q.append(s)
visit = [0] * (n + 1)
visit[s] = 1
while q:
i = q.popleft()
for j in G[i]:
if not visit[j]:
visit[j] = 1
q.append(j)
return visit
def bfs2(t):
q = deque()
q.append(t)
visit0 = [0] * (n + 1)
visit0[t] = 1
c = 0
while q:
i = q.popleft()
c += 1
for j in G[i]:
if not visit0[j] and not visit[j]:
visit0[j] = 1
q.append(j)
return c
def bfs3(t):
q = deque()
q.append(t)
visit[t] = 1
while q:
i = q.popleft()
for j in G[i]:
if not visit[j]:
visit[j] = 1
q.append(j)
return
n, m, s = map(int, input().split())
G = [[] for _ in range(n + 1)]
for _ in range(m):
u, v = map(int, input().split())
G[u].append(v)
visit = bfs1(s)
h = []
for i in range(1, n + 1):
if not visit[i]:
c = bfs2(i)
heapq.heappush(h, (-c, i))
ans = 0
while h:
c, i = heapq.heappop(h)
if not visit[i]:
ans += 1
bfs3(i)
print(ans)
``` | instruction | 0 | 93,246 | 1 | 186,492 |
Yes | output | 1 | 93,246 | 1 | 186,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
import sys
N = 5000 + 5
adj = [[] for i in range(N)]
mark = [0 for i in range(N)]
topo = []
sys.setrecursionlimit(6000)
#fin = open("999E.inp", "r")
#fou = open("999E.out", "w")
#n, m, s = map(int, fin.readline().split())
n, m, s = map(int, input().split())
for i in range(m):
#u, v = map(int, fin.readline().split())
u, v = map(int, input().split())
adj[u].append(v)
def topoSort(u):
mark[u] = 1
for j in range(len(adj[u])):
v = adj[u][j]
if (mark[v] == 0): topoSort(v)
topo.append(u)
def dfs(u):
mark[u] = 1
for j in range(len(adj[u])):
v = adj[u][j]
if (mark[v] == 0): dfs(v)
for i in range(1, n+1):
if (mark[i] == 0):
topoSort(i)
topo.reverse()
for i in range(1, n+1):
mark[i] = 0
dfs(s)
res = 0
for i in range(n):
v = topo[i]
if (mark[v] == 0):
res += 1
dfs(v)
#fou.write(str(res))
print(res)
``` | instruction | 0 | 93,247 | 1 | 186,494 |
Yes | output | 1 | 93,247 | 1 | 186,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
def parent(i,par,ind):
pp=i
if(par[i]==ind):
return 1
i=par[i]
while(1):
if(par[i]==ind):
return 1
if(par[i]==i):
par[i]=ind
break
i=par[i]
if(i==pp):
break
return 0
n,m,ind=[int(x) for x in input().split()]
par={}
for i in range(1,n+1):
par[i]=i
for i in range(m):
p,q=[int(x) for x in input().split()]
par[q]=p
c=0
for i in range(1,n+1):
#print(c,par)
if(i==ind):
continue
if(parent(i,par,ind)):
par[i]=ind
else:
c+=1
par[i]=ind
print(c)
``` | instruction | 0 | 93,248 | 1 | 186,496 |
No | output | 1 | 93,248 | 1 | 186,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
from collections import deque
import sys
input = sys.stdin.readline
def bfs1(s):
q = deque()
q.append(s)
group[s] = s
while q:
i = q.popleft()
for j in G[i]:
if not group[j]:
unite(s, j)
group[j] = s
q.append(j)
return
def bfs2(t):
q = deque()
q.append(t)
group[t] = t
while q:
i = q.popleft()
for j in G[i]:
if not group[j]:
unite(t, j)
group[j] = t
q.append(j)
elif not group[j] == s:
unite(i, j)
return
def get_root(s):
v = []
while not s == root[s]:
v.append(s)
s = root[s]
for i in v:
root[i] = s
return s
def unite(s, t):
rs, rt = get_root(s), get_root(t)
if not rs ^ rt:
return
if rank[s] == rank[t]:
rank[rs] += 1
if rank[s] >= rank[t]:
root[rt] = rs
size[rs] += size[rt]
else:
root[rs] = rt
size[rt] += size[rs]
return
def same(s, t):
return True if get_root(s) == get_root(t) else False
n, m, s = map(int, input().split())
G = [[] for _ in range(n + 1)]
for _ in range(m):
u, v = map(int, input().split())
G[u].append(v)
group = [0] * (n + 1)
root = [i for i in range(n + 1)]
rank = [1 for _ in range(n + 1)]
size = [1 for _ in range(n + 1)]
bfs1(s)
for i in range(2, n + 1):
if not group[i]:
bfs2(i)
ans = len(set([get_root(i) for i in range(1, n + 1)])) - 1
print(ans)
``` | instruction | 0 | 93,249 | 1 | 186,498 |
No | output | 1 | 93,249 | 1 | 186,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
n,m,s=map(int,input().split())
s-=1
edge=[[]for _ in range(n)]
to=[0]*n
for _ in range(m):
a,b=map(int,input().split())
a-=1
b-=1
to[b]+=1
edge[a].append(b)
to=[(to[i],i)for i in range(n)]
to.sort()
visited=[False]*n
visited[s]=True
queue=[s]
for node in queue:
for mode in edge[node]:
if visited[mode]:continue
visited[mode]=True
queue.append(mode)
ans=0
for _,node in to:
if visited[node]:continue
visited[node]=True
ans+=1
queue=[node]
for node in queue:
for mode in edge[node]:
if visited[mode]:continue
visited[mode]=True
queue.append(mode)
print(ans)
``` | instruction | 0 | 93,250 | 1 | 186,500 |
No | output | 1 | 93,250 | 1 | 186,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input
The first line of input consists of three integers n, m and s (1 โค n โค 5000, 0 โค m โค 5000, 1 โค s โค n) โ the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.
The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 โค u_i, v_i โค n, u_i โ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).
Output
Print one integer โ the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.
Examples
Input
9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1
Output
3
Input
5 4 5
1 2
2 3
3 4
4 1
Output
1
Note
The first example is illustrated by the following:
<image>
For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.
The second example is illustrated by the following:
<image>
In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def kosaraju(g):
# 1. For each vertex u of the graph, mark u as unvisited. Let l be empty.
size = len(g)
vis = [False]*size # vertexes that have been visited
l = [0]*size
T = size
t = [[] for _ in range(size)] # transpose graph
def visit(u):
nonlocal T, vis
stk = []
if not vis[u]: stk.append(u)
while stk:
x = stk[-1]
if not vis[x]:
vis[x] = True
for y in g[x]:
if not vis[y]:
stk.append(y)
t[y].append(x)
else:
stk.pop()
T -= 1
l[T] = x
# 2. For each vertex u of the graph do visit(u)
for u in range(len(g)):
visit(u)
c = [0]*size
def assign(u, root):
nonlocal vis
stk = [u]
while stk:
x = stk.pop()
if vis[x]:
vis[x] = False
c[x] = root
for y in t[x]:
stk.append(y)
# 3: For each element u of l in order, do assign(u, u)
for u in l:
assign(u, u)
return c
def condensed(g,s):
c = kosaraju(g)
size = len(g)
label = 0
new_c = [-1]*(size)
ref = dict()
for i in range(size):
if c[i] in ref:
new_c[i] = ref[c[i]]
else:
ref[c[i]] = label
new_c[i] = label
label += 1
new_g = [set() for _ in range(label)]
for i in range(size):
new_i = new_c[i]
for j in g[i]:
new_j = new_c[j]
if new_i != new_j:
new_g[new_i].add(new_j)
return new_g, new_c[s]
n,m,s = map(int,input().split())
G = [[] for _ in range(n)]
for _ in range(m):
x,y = map(int,input().split())
G[x-1].append(y-1)
G,s = condensed(G,s-1)
indeg = [0]*len(G)
for i in range(len(G)):
for j in G[i]:
indeg[j] += 1
ans = 0
for i in range(len(G)):
if indeg[i] == 0:
ans += 1
if indeg[s] == 0:
ans -= 1
print(ans)
``` | instruction | 0 | 93,251 | 1 | 186,502 |
No | output | 1 | 93,251 | 1 | 186,503 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,252 | 1 | 186,504 |
"Correct Solution:
```
class UnionFind:
def __init__(self, n):
self.g = [i for i in range(n)]
self.n = n
def whichclass(self, x):
if self.g[x]==x:
return x
else:
self.g[x] = self.whichclass(self.g[x])
return self.g[x]
def union(self, x, y):
cx = self.whichclass(x)
cy = self.whichclass(y)
if cx==cy:
return
minc = min(cx, cy)
maxc = max(cx, cy)
self.g[maxc] = minc
def __getitem__(self, x):
return self.whichclass(x)
N, M = map(int, input().split())
UF = UnionFind(N)
for i in range(M):
a, b = map(int, input().split())
a, b = a-1, b-1
UF.union(a, b)
count = [0 for i in range(N)]
for i in range(N):
count[UF[i]] = 1
print(sum(count)-1)
``` | output | 1 | 93,252 | 1 | 186,505 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,253 | 1 | 186,506 |
"Correct Solution:
```
N, M = map(int, input().split())
F = [[] for _ in range(N)]
P = [0] * N
for i in range(M):
A, B = map(int, input().split())
F[A-1].append(B-1)
F[B-1].append(A-1)
c=0
l = []
for i in range(N):
if P[i] == 0:
c+=1
l.append(i)
while len(l)>0:
p=l.pop()
P[p] = c
for j in F[p]:
if P[j] == 0:
l.append(j)
print(c-1)
``` | output | 1 | 93,253 | 1 | 186,507 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,254 | 1 | 186,508 |
"Correct Solution:
```
N,M=map(int,input().split())
road=[[] for _ in range(N)]
for i in range(M):
a,b=map(int,input().split())
road[a-1].append(b-1)
road[b-1].append(a-1)
ischecked=[0]*(N)
ans=0
for i in range(N):
if ischecked[i]==1:
continue
ischecked[i]=1
q=[i]
while q:
now=q.pop()
nex=road[now]
for j in nex:
if ischecked[j]!=0:
continue
q.append(j)
ischecked[j]=1
ans+=1
print(ans-1)
``` | output | 1 | 93,254 | 1 | 186,509 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,255 | 1 | 186,510 |
"Correct Solution:
```
N, M = map(int, input().split())
cla = [i for i in range(N)]
count = set()
def root(a):
change = []
while cla[a] != a:
change.append(a)
a = cla[a]
for i in change:
cla[i] = a
return a
for _ in range(M):
a, b = map(int, input().split())
a -= 1
b -= 1
if root(a) != root(b):
cla[root(b)] = root(a)
for i in range(N):
count |= {root(i)}
print(len(count) - 1)
``` | output | 1 | 93,255 | 1 | 186,511 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,256 | 1 | 186,512 |
"Correct Solution:
```
N, M = map(int, input().split())
P = [-1 for i in range(N)]
def par(a):
L = []
while P[a] >= 0:
L.append(a)
a = P[a]
for l in L:
P[l] = a
return a
def unite(a, b):
if par(a) != par(b):
if P[par(b)] >= P[par(a)]:
if P[par(b)] == P[par(a)]: P[par(a)] -= 1
P[par(b)] = par(a)
else:
P[par(a)] = par(b)
ans = N - 1
for _ in range(M):
a, b = map(int, input().split())
a, b = a-1, b-1
if par(a) == par(b): continue
unite(a, b)
ans -= 1
print(ans)
``` | output | 1 | 93,256 | 1 | 186,513 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,257 | 1 | 186,514 |
"Correct Solution:
```
N,M = map(int,input().split())
par = [0]+[-1 for i in range(N)]
def find(x):
if par[x] == -1:
return x
else:
par[x] = find(par[x]) #็ต่ทฏๅง็ธฎ
return par[x]
def same(x,y):
return find(x) == find(y)
def unite(x,y):
x = find(x)
y = find(y)
if x == y:
return 0
par[x] = y
for i in range(M):
a,b = map(int,input().split())
unite(a,b)
print(len([i for i in par if i == -1])-1)
``` | output | 1 | 93,257 | 1 | 186,515 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,258 | 1 | 186,516 |
"Correct Solution:
```
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
def unite(x,y):
if same(x,y)==False:
rx=root(x);ry=root(y)
if rank[rx]>rank[ry]:
parent[ry]=rx
size[rx]+=size[ry]
else:
parent[rx]=ry
size[ry]+=size[rx]
if rank[rx]==rank[ry]:
rank[ry]+=1
def root(x):
if x!=parent[x]:
return root(parent[x])
return x
def same(x,y):
return root(x)==root(y)
parent=list(range(n))
rank=[0]*n
size=[1]*n
for i in range(m):
a,b=map(int,input().split())
unite(a-1,b-1)
kind=set()
for i in range(n):
r=parent[root(i)]
kind.add(r)
print(len(kind)-1)
``` | output | 1 | 93,258 | 1 | 186,517 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1 | instruction | 0 | 93,259 | 1 | 186,518 |
"Correct Solution:
```
import sys
class UnionFind():
def __init__(self, n):
self.parents = [-1] * n
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
N, M = map(int, input().split())
uf = UnionFind(N)
for _ in range(M):
a, b = map(lambda x: int(x) - 1, sys.stdin.readline().split())
uf.union(a, b)
print(sum(p < 0 for p in uf.parents) - 1)
``` | output | 1 | 93,259 | 1 | 186,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
from collections import deque
N, M = map(int, input().split())
AB = [[] for _ in range(N)]
for _ in range(M):
a,b = map(lambda x: int(x)-1, input().split())
AB[a].append(b)
AB[b].append(a)
visited = [0] * N
ans = []
for i in range(N):
if visited[i]:
continue
group = [i]
d = deque(AB[i])
while d:
s = d.popleft()
if visited[s]:
continue
visited[s] = 1
group.append(s)
for k in AB[s]:
d.append(k)
ans.append(len(group))
print(len(ans)-1)
``` | instruction | 0 | 93,260 | 1 | 186,520 |
Yes | output | 1 | 93,260 | 1 | 186,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
def find(x):
if par[x] < 0:
return x
else:
par[x] = find(par[x])
return par[x]
def same(x, y):
return find(x) == find(y)
def unite(x, y):
x = find(x)
y = find(y)
if x == y:
return
if par[x] > par[y]:
x, y = y, x
par[x] += par[y]
par[y] = x
N, M = map(int, input().split())
par = [-1 for i in range(N)]
for i in range(M):
a, b = map(int, input().split())
unite(a - 1, b - 1)
cnt = -1
num = {0}
for i in par:
if i <= -1:
cnt += 1
print(cnt)
``` | instruction | 0 | 93,261 | 1 | 186,522 |
Yes | output | 1 | 93,261 | 1 | 186,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
#Union Find
#xใฎๆ นใๆฑใใ
def find(x):
if par[x] < 0:
return x
else:
tank = []
while par[x] >= 0:
tank.append(x)
x = par[x]
for elt in tank:
par[elt] = x
return x
#xใจyใฎๅฑใใ้ๅใฎไฝตๅ
def unite(x,y):
x = find(x)
y = find(y)
if x == y:
return False
else:
#sizeใฎๅคงใใใปใใx
if par[x] > par[y]:
x,y = y,x
par[x] += par[y]
par[y] = x
return True
n,m = map(int, input().split())
par = [-1] * n
for i in range(m):
a,b = map(int,input().split())
unite(a-1,b-1)
ans = set([])
for i in range(n):
ans.add(find(i))
print(len(ans)-1)
``` | instruction | 0 | 93,262 | 1 | 186,524 |
Yes | output | 1 | 93,262 | 1 | 186,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
n,m=map(int,input().split())
par=[i for i in range(n)]
size=[1]*n
def find(x):
if par[x]==x:
return x
else:
par[x]=find(par[x])
return par[x]
def union(a,b):
x=find(a)
y=find(b)
if x!=y:
if size[x]<size[y]:
par[x]=par[y]
size[y]+=size[x]
else:
par[y]=par[x]
size[x]+=size[y]
else:
return
for i in range(m):
a,b=map(int,input().split())
union(a-1,b-1)
ans={find(i) for i in range(n)}
print(len(ans)-1)
``` | instruction | 0 | 93,263 | 1 | 186,526 |
Yes | output | 1 | 93,263 | 1 | 186,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
N, M = (int(i) for i in input().split())
AB = [input().split() for i in range(M)]
route = set()
for ab in AB:
route.add(ab[0])
route.add(ab[1])
if(len(route) == N):
break
print(N - len(route))
``` | instruction | 0 | 93,264 | 1 | 186,528 |
No | output | 1 | 93,264 | 1 | 186,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
from typing import List, Dict
MAX_N = 10 ** 5
visited = [False] * MAX_N
cluster = [0] * MAX_N
edges: Dict[int, List[int]] = {}
def dfs(node: int, cluster_num: int) -> None:
cluster[node] = cluster_num
visited[node] = True
if node not in edges:
return
for v in edges[node]:
if visited[v]:
continue
dfs(v, cluster_num)
def connect_cities(node_num: int, edge_num: int) -> int:
i = 0
for s in range(node_num):
if visited[s]:
continue
dfs(s, i)
i += 1
return i - 1
if __name__ == "__main__":
N, M = map(int, input().split())
for _ in range(M):
a, b = map(int, input().split())
a, b = a - 1, b - 1
edges.setdefault(a, [])
edges[a].append(b)
edges.setdefault(b, [])
edges[b].append(a)
ans = connect_cities(N, M)
print(ans)
``` | instruction | 0 | 93,265 | 1 | 186,530 |
No | output | 1 | 93,265 | 1 | 186,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
def s(way):
global w
global wc
for i in way:
c[i] = 1
if wc[i] == 0:
wc[i] =1
for j in w[i]:
c[j]=1
s(w[j])
n,m = map(int,input().split())
a = [0 for i in range(m)]
b = [0 for i in range(m)]
w = [[] for i in range(n)]
wc = [0 for i in range(n)]
c = [0 for i in range(n)]
for i in range(m):
a[i],b[i] = map(int,input().split())
w[a[i]-1].append(b[i]-1)
w[b[i]-1].append(a[i]-1)
ans =-1
for i in range(n):
if c[i] == 0:
c[i] = 1
wc[i] = 1
ans += 1
s(w[i])
print(ans)
``` | instruction | 0 | 93,266 | 1 | 186,532 |
No | output | 1 | 93,266 | 1 | 186,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.
Snuke can perform the following operation zero or more times:
* Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.
After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).
What is the minimum number of roads he must build to achieve the goal?
Constraints
* 2 \leq N \leq 100,000
* 1 \leq M \leq 100,000
* 1 \leq A_i < B_i \leq N
* No two roads connect the same pair of cities.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
:
A_M B_M
Output
Print the answer.
Example
Input
3 1
1 2
Output
1
Submitted Solution:
```
n,m=map(int,input().split())
if n>m:
print(n-1-m)
else:
print(0)
``` | instruction | 0 | 93,267 | 1 | 186,534 |
No | output | 1 | 93,267 | 1 | 186,535 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,391 | 1 | 186,782 |
"Correct Solution:
```
import sys,heapq
from collections import defaultdict,deque
input = sys.stdin.readline
n,m = map(int,input().split())
#+่ทฏ็ท*chg : ใใฎ่ทฏ็ทใใผใ
chg = 10**6
edge=defaultdict(set)
used =defaultdict(bool)
for i in range(m):
p,q,c = map(int,input().split())
edge[p].add(p+c*chg)
edge[p+c*chg].add(p)
edge[p+c*chg].add(q+c*chg)
edge[q+c*chg].add(p+c*chg)
edge[q].add(q+c*chg)
edge[q+c*chg].add(q)
used[p] = False
used[q] = False
used[p+c*chg] = False
used[q+c*chg] = False
edgelist = deque()
edgelist.append((1,0))
res = float("inf")
while len(edgelist):
x,cost = edgelist.pop()
used[x] = True
if x == n:
res = cost
break
for e in edge[x]:
if used[e]:
continue
#ใใผใ โๆนๆญ
if x <= 10**5 and chg <= e:
edgelist.appendleft((e,cost+1))
else:
edgelist.append((e,cost))
if res == float("inf"):
print(-1)
else:
print(res)
``` | output | 1 | 93,391 | 1 | 186,783 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,392 | 1 | 186,784 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import defaultdict
"""
(้ง
ใไผ็คพ)ใ้ ็นใซใฐใฉใใๆใคใ้ ็นๆฐO(M)ใ
ใใฎใพใพ่พบใ่ฒผใใจ่พบใๅคใใชใใใใใ
(้ง
ใไผ็คพ) -> (้ง
ใ็กๅฑๆง) -> (้ง
ใไผ็คพ)
"""
L = 32
mask = (1 << L) - 1
N,M = map(int,input().split())
graph = defaultdict(list)
for _ in range(M):
p,q,c = map(int,input().split())
p <<= L
q <<= L
graph[p].append(p+c)
graph[p+c].append(p)
graph[q].append(q+c)
graph[q+c].append(q)
graph[p+c].append(q+c)
graph[q+c].append(p+c)
INF = 10 ** 9
dist = defaultdict(lambda: INF)
start = 1 << L
goal = N << L
q = [start] # 0 ใไผ็คพใซๅฑใใฆใใชใ็ถๆ
dist[start] = 0
d = 0
while q:
qq = []
while q:
x = q.pop()
if x & mask == 0:
for y in graph[x]:
if dist[y] <= d + 1:
continue
dist[y] = d + 1
qq.append(y)
else:
for y in graph[x]:
if dist[y] <= d:
continue
dist[y] = d
q.append(y)
d += 1
q = qq
answer = dist[goal]
if answer == INF:
answer = -1
print(answer)
``` | output | 1 | 93,392 | 1 | 186,785 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,393 | 1 | 186,786 |
"Correct Solution:
```
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
from collections import deque
def solve():
dist = [inf] * len(to)
q = deque()
q.append((0, 0))
dist[0] = 0
while q:
d, i = q.popleft()
if d > dist[i]: continue
if i < n:
for j in to[i]:
if dist[j] <= d + 1: continue
dist[j] = d + 1
q.append((d + 1, j))
else:
for j in to[i]:
if j == n - 1:
print(dist[i])
exit()
if dist[j] <= d: continue
dist[j] = d
q.appendleft((d, j))
print(-1)
inf=10**9
n,m=MI()
to=[]
uctoi={}
for u in range(n):
to.append([])
uctoi[u,0]=u
for _ in range(m):
u,v,c=MI()
u,v=u-1,v-1
if (u,c) not in uctoi:
i=uctoi[u,c]=len(to)
to.append([])
to[i].append(u)
to[u].append(i)
else:i=uctoi[u,c]
if (v,c) not in uctoi:
j=uctoi[v,c]=len(to)
to.append([])
to[j].append(v)
to[v].append(j)
else:j=uctoi[v,c]
to[i].append(j)
to[j].append(i)
#print(to)
#print(fin)
solve()
``` | output | 1 | 93,393 | 1 | 186,787 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,394 | 1 | 186,788 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
def bfs01(g, s):
d = {}
for v in g.keys():
d[v] = sys.maxsize
d[s] = 0
if s not in d:
return d
q = deque()
q.append((d[s], s))
NM = 1 << 30
while q:
_, u = q.popleft()
for v in g[u]:
if v > NM:
c = 0
else:
c = 1
alt = c + d[u]
if d[v] > alt:
d[v] = alt
if c == 0:
q.appendleft((d[v], v))
else:
q.append((d[v], v))
return d
@mt
def slv(N, M):
g = defaultdict(list)
NM = 30
for _ in range(M):
p, q, c = input().split()
p = int(p)
q = int(q)
c = int(c)
cp = (c << NM) + p
cq = (c << NM) + q
g[cp].append(cq)
g[cq].append(cp)
g[cp].append(p)
g[cq].append(q)
g[p].append((cp))
g[q].append((cq))
d = bfs01(g, 1)
return -1 if N not in d or d[N] == sys.maxsize else d[N]
def main():
N, M = read_int_n()
print(slv(N, M))
if __name__ == '__main__':
main()
``` | output | 1 | 93,394 | 1 | 186,789 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,395 | 1 | 186,790 |
"Correct Solution:
```
from collections import deque,defaultdict
import sys
input = sys.stdin.readline
N,M=map(int,input().split())
mod =10**7
table=defaultdict(set)
visit=defaultdict(int)
for i in range(M):
a,b,c=map(int,input().split())
table[a].add(a+c*mod)
table[b].add(b+c*mod)
table[a+c*mod].add(a)
table[b+c*mod].add(b)
table[a+c*mod].add(b+c*mod)
table[b+c*mod].add(a+c*mod)
visit[a]=0
visit[b]=0
visit[b+c*mod]=0
visit[a+c*mod]=0
H=deque()
H.append((1,0))
visit[1]=1
ans=mod
while H:
#print(H)
x,cost=H.popleft()
visit[x] = 1
if x==N:
ans=min(ans,cost)
break
for y in table[x]:
if visit[y]==1:
continue
if x<=10**5 and mod<=y:
H.append((y,cost+1))
else:
H.appendleft((y,cost))
if ans==mod:
print(-1)
else:
print(ans)
``` | output | 1 | 93,395 | 1 | 186,791 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,396 | 1 | 186,792 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
import sys
readline = sys.stdin.readline
read = sys.stdin.read
n,m = map(int, readline().split())
g = {}
for _ in range(m):
p,q,c = map(int, readline().split())
pc = ((p-1)<<20)+c
qc = ((q-1)<<20)+c
pp = (p-1)<<20
qq = (q-1)<<20
if pc not in g: g[pc] = []
if qc not in g: g[qc] = []
if pp not in g: g[pp] = []
if qq not in g: g[qq] = []
g[pc].append(qc)
g[pc].append(pp)
g[qc].append(pc)
g[qc].append(qq)
g[pp].append(pc)
g[qq].append(qc)
if 0 not in g: g[0] = []
from collections import deque
q = deque([(0,0)])
res = {0:0}
mask = (1<<20)-1
while q:
vl,dv = q.popleft()
if res[vl] < dv: continue
if (vl>>20) == n-1:
res[(n-1)<<20] = dv+1
break
for tl in g[vl]:
ndv = dv + (vl&mask==0 or tl&mask==0)
if tl not in res or res[tl] > ndv:
res[tl] = ndv
if vl&mask==0 or tl&mask==0:
q.append((tl,ndv))
else:
q.appendleft((tl,ndv))
if (n-1)<<20 in res:
print(res[(n-1)<<20]//2)
else:
print(-1)
``` | output | 1 | 93,396 | 1 | 186,793 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,397 | 1 | 186,794 |
"Correct Solution:
```
import sys
from collections import deque, defaultdict
def bfs01(s, t, links):
q = deque([(0, s, -1)]) # cost, station, last-company
visited = set()
while q:
d, v, e = q.popleft()
if v == t:
return d
if (v, e) in visited:
continue
visited.add((v, e))
if e == 0:
lv = links[v]
for c in lv:
for u in lv[c]:
if (u, c) in visited:
continue
q.append((d + 1, u, c))
else:
for u in links[v][e]:
if (u, e) in visited:
continue
q.appendleft((d, u, e))
if (v, 0) not in visited:
q.appendleft((d, v, 0))
return -1
readline = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
n, m = map(int, readline().split())
links = [defaultdict(set) for _ in range(n)]
pqc = list(map(int, read().split()))
for p, q, c in zip(pqc[0::3], pqc[1::3], pqc[2::3]):
p -= 1
q -= 1
links[p][c].add(q)
links[q][c].add(p)
print(bfs01(0, n - 1, links))
``` | output | 1 | 93,397 | 1 | 186,795 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1 | instruction | 0 | 93,398 | 1 | 186,796 |
"Correct Solution:
```
from collections import defaultdict,deque
import sys
sys.setrecursionlimit(10**6)
input=sys.stdin.readline
N,M=map(int,input().split())
if M==0:
print(-1)
exit()
G=defaultdict(set)
for i in range(M):
a,b,c=map(int,input().split())
G[a+(c<<30)].add(b+(c<<30))
G[b+(c<<30)].add(a+(c<<30))
G[a].add(a+(c<<30))
G[b].add(b+(c<<30))
G[a+(c<<30)].add(a)
G[b+(c<<30)].add(b)
def BFS(x):
d={}
for k in G.keys():
d[k]=float("inf")
stack=deque([(x,0)])
if x not in d:
return -1
while stack:
s,m=stack.popleft()
if s==N:
return m
if d[s]>m:
d[s]=m
if s>=(1<<30):
for i in G[s]:
if d[i]>10**30:
stack.appendleft((i,m))
else:
for i in G[s]:
if d[i]>10**30:
stack.append((i,m+1))
return -1
print(BFS(1))
``` | output | 1 | 93,398 | 1 | 186,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2ร10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
def main():
N,M=map(int,input().split())
if M == 0:
print(-1)
exit()
INF = 10**6
pqc = [tuple(map(lambda x:int(x)-1,input().split())) for _ in range(M)]
to1d={v:{} for v in range(N)}
count=2
for p,q,c in pqc:
if not to1d[p].get(c):
to1d[p][c] = count
count += 1
if not to1d[q].get(c):
to1d[q][c] = count
count += 1
G = [{} for _ in range(N+count)]
for p,q,c in pqc:
v1,v2 = to1d[p][c],to1d[q][c]
G[v1][v2] = G[v2][v1] = 0
for i in range(N):
if len(to1d[i])<=1: continue
for c in to1d[i].keys():
v = to1d[i][c]
G[v][count] = 1
G[count][v] = 0
count += 1
def dijkstra(start = 0, goal = 1):
from heapq import heappop, heappush
NN = len(G)
d = [INF]*NN
d[start] = 0
que = []
heappush(que, start)
while que:
p = divmod(heappop(que),NN)
v = p[1]
if d[v] < p[0]: continue
for u in G[v].keys():
if d[u] > d[v] + G[v][u]:
d[u] = d[v] + G[v][u]
heappush(que, d[u]*NN+u)
if v == goal: return d[goal]
return d[goal]
for c in to1d[0].keys():
v = to1d[0][c]
G[0][v] = 1
for c in to1d[N-1].keys():
v = to1d[N-1][c]
G[v][1] = 0
ans = dijkstra(0,1)
print(ans if ans < INF else -1)
if __name__=="__main__":
main()
``` | instruction | 0 | 93,399 | 1 | 186,798 |
Yes | output | 1 | 93,399 | 1 | 186,799 |
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