hdoj 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): 4662    Accepted Submission(s): 1833

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

25001001000001001111011100050111100000010000110001110

 

Sample Output

Case #1: YesCase #2: No

 

判断是否有向图成环，用拓扑排序

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;int map[2010][2010];int in[2010];int n,k;void topo(){    int ans=0;    queue<int> q;    for(int i=0;i<n;i++)    {        if(in[i]==0)            q.push(i);    }    while(!q.empty())    {        int a=q.front();        in[a]--;        ans++;        q.pop();        for(int j=0;j<n;j++)        {            if(map[a][j])            {                in[j]--;                if(in[j]==0)                    q.push(j);            }        }    }    if(ans!=n)        printf("Case #%d: Yes\n",k++);    else        printf("Case #%d: No\n",k++);}int main(){    int t,x;    char str[2010];    k=1;    scanf("%d",&t);    while(t--)    {        memset(map,0,sizeof(map));        memset(in,0,sizeof(in));        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%s",str);            for(int j=0;j<n;j++)            {                            if(str[j]=='1')                {                    map[i][j]=1;                    in[j]++;                }            }        }        topo();    }    return 0;}