hdoj 4324 Triangle LOVE 【拓扑排序判断是否存在可行解】

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3509 Accepted Submission(s): 1364

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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.

It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,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

三角恋关系：A喜欢B，B喜欢C，C喜欢A。

大致题意：给你一个N*N的矩阵，A[i][j] = 1表示第i个人喜欢第j个人，为0时表示不喜欢。问你N个人里面是否存在三角恋关系。

思路：直接正向拓扑判断可行解即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define MAXN 2000+10
using namespace std;
vector<int> G[MAXN];
int in[MAXN];
int N;
int k = 1;
void init()
{
    for(int i = 1; i <= N; i++)
        G[i].clear(), in[i] = 0;
}
void getMap()
{
    char str[2100];
    for(int i = 1; i <= N; i++)
    {
        scanf("%s", str);
        for(int j = 0; j < N; j++)
        {
            if(str[j] > '0')
                G[i].push_back(j+1), in[j+1]++;
        }
    }
}
void solve()
{
    queue<int> Q;
    for(int i = 1; i <= N; i++)
        if(in[i] == 0) Q.push(i);
    int num = 0;
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        num++;
        for(int i = 0; i < G[u].size(); i++)
        {
            int v = G[u][i];
            if(--in[v] == 0)
                Q.push(v);
        }
    }
    printf("Case #%d: ", k++);
    if(num != N)//存在
        printf("Yes\n");
    else
        printf("No\n");
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        init();
        getMap();
        solve();
    }
    return 0;
}