Triangle LOVE 4324 （拓扑排序）

Triangle LOVE

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 12   Accepted Submission(s) : 7
[align=left]Problem Description[/align]
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”.

 

[align=left]Input[/align]
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[sub]i,j[/sub] = 1 means i-th people loves j-th people, otherwise A[sub]i,j[/sub] = 0. It is guaranteed that the given relationship is a tournament, that is, A[sub]i,i[/sub]= 0, A[sub]i,j[/sub] ≠ A[sub]j,i[/sub](1<=i, j<=n,i≠j).
 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

 

[align=left]Sample Output[/align]

Case #1: Yes
Case #2: No

/*
题意：：
有T组数据，每组数据第一行为N，接下来是N*N一个字符数组，若为‘1’表示喜欢
‘0’表示不喜欢。如：：
5
00100  //表示第1个人喜欢第3个。
10000  //第2个人喜欢第1个。
01001  //第3个人喜欢2和5.
11101  //第4人喜欢1,2,3,5
11000  //第5人喜欢1，2
判断是否产生三角恋关系或多角恋关系（是否成环）
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[2010][2010];
int in[2010];
int n;
int topo(int n)
{
int i,j;
for(j=0;j<n;j++)
{
int m=-1;
for(i=0;i<n;i++)
{
if(in[i]==0)
{
in[i]--;
m=i;
break;
}
}
if(m==-1)//如果成环，直接跳出循环
break;
for(i=0;i<n;i++)//否则将他们逐个剔除
{
if(map[m][i]=='1')
{
in[i]--;
map[m][i]='0';
}
}
}
if(j<n)
printf("Yes\n");
else
printf("No\n");
}
int main(){
int t,s=1;
scanf("%d",&t);
getchar();
while(t--)
{
int i,j;
scanf("%d",&n);
memset(in,0,sizeof(in));
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
for(j=0;j<n;j++)
{
if(map[i][j]=='1')
{
in[j]++;
}
}
}
printf("Case #%d: ",s++);
topo(n);
}
return 0;
}