HDU 4324 Triangle LOVE 【判环+拓扑】

Time limit

1000 ms

Memory limit

65536 kB

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, Ai,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

思路：由题意可以知道，对于任意给定的一个有环图，一定存在着三元环（想想这是为啥？）
假设有环上三个相邻的点a-> b-> c
如果c->a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元环递归即可
拓扑解决

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int maxn =2e3+10;
char map[maxn][maxn];
int indeg[maxn],n;
vector<int>G[maxn]; //邻接表存图
queue<int>q;
void topsort()
{
    int num=0;
    while(!q.empty())
        q.pop();
    for(int i=0;i<n;i++)  //将入度为0的点加入队列
    {
        if(!indeg[i])
            q.push(i);
    }
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        num++;
        //这一步是判断与now相连的边在入度-1的情况下是否为0
        for(int i=0;i<G[now].size();i++) 
        {
            if(--indeg[G[now][i]]==0)
                q.push(G[now][i]);  //入度为0的话进入队列
        }
    }
    if(num!=n)
        printf("Yes\n");
    else
        printf("No\n");
}
int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(indeg,0,sizeof(indeg)); //入度初始化为0
        for(int i=0;i<n;i++)  //清空
            G[i].clear();
        for(int i=0;i<n;i++)
        {
            getchar();
            scanf("%s",map[i]);
            for(int j=0;j<n;j++)
            {
                if(map[i][j]=='1')
                {
                    G[i].push_back(j);  
                    indeg[j]++;    //i-->j有一条边，j的入度加一
                }
            }
        }
        printf("Case #%d: ",cas++);
        topsort();
    }
    return 0;
}

HDU 4324 Triangle LOVE 【判环+拓扑】

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