HDU 4324 Triangle LOVE （拓扑排序）

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1707    Accepted Submission(s): 729

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

 

Recommend

zhoujiaqi2010

 

题意分析(转载)：
此题可以一遍拓扑排序判环求解 即只需要找到一个环，
就必定存在三元环 证明如下： 假设存在一个n元环，
因为a->b有边，b->a必定没边，反之也成立
所以假设有环上三个相邻的点a-> b-> c,那么如果c->a间有边，
就已经形成了一个三元环，如果c->a没边，那么a->c肯定有边，
这样就形成了一个n-1元环。。。。
所以只需证明n大于3时一定有三元环即可，显然成立。

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

using namespace std;

const int N=;

int n,indeg[N];     //存储的是节点的入度
char str[N][N];

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int flag=;
        memset(indeg,,sizeof(indeg));  //将所有的节点入度初始化为0
        int i,j;
        for(i=;i<n;i++){
            scanf("%s",str[i]);
            for(j=;j<n;j++)
                if(str[i][j]=='')  //如果i喜欢j,则把j的入度加1
                    indeg[j]++;
        }
        for(i=;i<n;i++){
            for(j=;j<n;j++)
                if(indeg[j]==)     //找出入度为0的节点
                    break;
            if(j==n){       //任何一个节点的入度都不为0，说明存在环了，则必有三角恋
                flag=;
                break;
            }else{
                indeg[j]--; //除去当前结点
                for(int k=;k<n;k++)    //把从这个节点出发的引起的节点的入度都减去1
                    if(str[j][k]=='')
                        indeg[k]--;
            }
        }
        if(flag)
            printf("Case #%d: Yes\n",++cases);
        else
            printf("Case #%d: No\n",++cases);
    }
    return ;
}