4324 Triangle LOVE

Triangle LOVE

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

Total Submission(s): 3543    Accepted Submission(s): 1382

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

简单的拓扑排序，这个题出的有问题，只需要判断是否出现环就可以了，不需要判断是三元还是多元环.....

也就是在所有的元素都未遍历完全的时候，所有的顶点的入度都不为零，这样肯定形成了自环，自己理解的还不够，慢慢学习吧，另外这个题的输入特别坑，单个单个的字符输入会超时，需要每次输入一个字符串，才可以正常通过....晕死....

#include<stdio.h>#include<string.h>int n,x[2005][2005],ind[2005],cnt=0;void tpsort(){    int i,j,k;     for(i=0;i<n;++i)    {        for(j=0;j<n;++j)        {            if(ind[j]==0)            {                break;            }        }        if(j>n-1)//程序内部出现所有的节点都不为零...        {            printf("Case #%d: Yes\n",++cnt);            return;        }        ind[j]=-1;        for(k=0;k<n;++k)        {            if(x[j][k])            {                --ind[k];            }        }    }    printf("Case #%d: No\n",++cnt);}int main(){    int t,i,j,b;    char a,s[2005];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(x,0,sizeof(x));        memset(ind,0,sizeof(ind));        for(i=0;i<n;++i)        {            scanf("%s",s);            for(j=0;s[j]!=0;++j)            {                if(s[j]=='1'&&!x[i][j])                {                    x[i][j]=1;                    ++ind[j];                }            }        }        tpsort();    }    return 0; }