Codeforces 117C. Cycle 寻找环

C. Cycle

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes u and v (u ≠ v) exists either an edge going from u to v, or an edge from v to u.

You are given a tournament consisting of n vertexes. Your task is to find there a cycle of length three.

Input

The first line contains an integer n (1 ≤ n ≤ 5000). Next n lines contain the adjacency matrix A of the graph (without spaces). A_i, j = 1 if the graph has an edge going from vertex i to vertex j, otherwise A_i, j = 0. A_i, j stands for the j-th character in the i-th line.

It is guaranteed that the given graph is a tournament, that is, A_i, i = 0, A_i, j ≠ A_j, i (1 ≤ i, j ≤ n, i ≠ j).

Output

Print three distinct vertexes of the graph a₁, a₂, a₃ (1 ≤ a_i ≤ n), such that A_a1, a2 = A_a2, a3 = A_a3, a1 = 1, or "-1", if a cycle whose length equals three does not exist.

If there are several solutions, print any of them.

Sample test(s)

input

5
00100
10000
01001
11101
11000

output

1 3 2 

input

5
01111
00000
01000
01100
01110

output

-1

找到并输出一条长度为3的环。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
bool v[5555]={0};
int n;
char s[5555][5555];
bool dfs(int i,int dad)
{
    v[i]=true;
    for (int j=1;j<=n;j++)
    {
        if (s[i][j]-'0')
        {
            if (s[j][dad]-'0')
            {
                printf("%d %d %d\n",dad,i,j);
                return true;
            }
            if (!v[j])
            {
                if (dfs(j,i)) return true;
            }
        }
    }
    return false;
}

int main()
{
    cin>>n;

    for (int i=1;i<=n;i++)
    {
        scanf("%s",s[i]+1);
    }

    for (int i=1;i<=n;i++)
    {
        if (!v[i])
        {
            if (dfs(i,i)) return 0;
        }
    }
    printf("-1\n");
    return 0;
}