hdu - 4324 Triangle LOVE【dfs】

 

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4610    Accepted Submission(s): 1809

Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

 

 

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,i(1<=i, j<=n,i≠j).

 

 

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.

 

 

Sample Input

2 5 00100 10000 01001 11101 11000 5 01111 00000 01000 01100 01110

 

 

Sample Output

Case #1: Yes Case #2: No

 

 

Author

BJTU

 

 

Source

2012 Multi-University Training Contest 3

每次进行3次DFS， 注意标记就可以了。

 

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> PAI;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 2000 + 10;
char data[MAXN][MAXN];
bool vis[MAXN];
int n;
vector<int> G[MAXN];
int Init() {
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < n; i++) G[i].clear();
}
bool dfs(int x, int step, int init) {
    if (step == 3 && x == init) return true;
    if (step == 3) return false;
    for (int i = 0; i < G[x].size(); i++) {
        if (vis[G[x][i]]) return false;
        if (dfs(G[x][i], step + 1, init)) return true;
        vis[x] = true;
    }
    return false;
}
int main() {
    int T;
    scanf("%d", &T);
    int Kcase = 0;
    while (T--) {
        bool flag = false;
        scanf("%d", &n); Init();
        for (int i = 0; i < n; i++) {
            scanf("%s", data[i]);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (data[i][j] == '1') G[i].push_back(j);
            }
        }
        for (int i = 0; i < n; i++) {
            if (dfs(i, 0, i)) {flag = true; break;}
        }
        if (flag) printf("Case #%d: Yes\n", ++Kcase);
        else printf("Case #%d: No\n", ++Kcase);
    }
    return 0;
}