CodeForce 117C Cycle DFS

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.

Examples

Input

Output

1 3 2

Input

Output

-1

OJ-ID： 
CodeForce 117C

author：
Caution_X

date of submission：
20190930

tags：
DFS

description modelling：
给定一个有向图，边权都为1，问能否找到权值和为3的环，找到则输出对应的点标号，否则输出-1

major steps to solve it：
1.vis[]表示该点是否访问过
2.从一个未被访问过的点开始DFS，找到与该点相连且未被访问过的点继续DFS
3.如果形成了环，结束DFS，否则继续2操作

AC CODE:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <iomanip>
 
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")
#define maxn 5005
#define MOD 1000000007
#define mem(a , b) memset(a , b , sizeof(a))
#define LL long long
#define ULL unsigned long long
#define FOR(i , n) for(int i = 1 ;  i<= n ; i ++)
typedef pair<int , int> pii;
const long long INF= 0x3fffffff;
int n , flag;
int a , b , c;
char arr[maxn][maxn];
int vis[maxn];
 
void dfs(int u , int v)
{
    if(a && b && c) return;
    vis[u] = 1;
    for(int i = 0 ; i < n &&(!a ||!b || !c); i ++)
    {
        if(arr[u][i] == '1' )
        {
            if(v != -1 && arr[i][v] == '1')
            {
                a = v + 1 , b = u + 1 , c = i + 1;
                return ;
            }
            if(!vis[i]) dfs(i , u);
        }
    }
 
}
 
int main()
{
    while(scanf("%d" , &n) != EOF)
    {
        mem(vis , 0);
        for(int i = 0 ; i < n; i ++)
        {
            scanf("%s" , arr[i]);
        }
        flag = 0;
        a = b = c  = 0;
        for(int i = 0 ; i < n ; i ++)
        {
            if(!vis[i])
            {
                dfs(i , -1);
            }
        }
        if(!a) printf("-1\n");
        else printf("%d %d %d\n" , a , b , c);
    }
    return 0;
}

View Code

原文地址：https://www.cnblogs.com/cautx/p/11612451.html