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). Ai, j = 1 if the graph has an edge going from vertex i to vertex j, otherwise Ai, j = 0. Ai, j stands for the j-th character in the i-th line.
It is guaranteed that the given graph is a tournament, that is, Ai, i = 0, Ai, j ≠ Aj, i (1 ≤ i, j ≤ n, i ≠ j).
Output
Print three distinct vertexes of the graph a1, a2, a3 (1 ≤ ai ≤ n), such that Aa1, a2 = Aa2, a3 = Aa3, 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
50010010000010011110111000
Output
1 3 2
Input
50111100000010000110001110
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DFS3.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;
}
原文地址：https://www.cnblogs.com/cautx/p/11612451.html