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): 3548    Accepted Submission(s): 1384


[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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.

It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(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

判断是否存在回路即可。

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define INF 0x7fffffff
#define MAXN 2001
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
using namespace std;
int pri[MAXN][MAXN];
int v[MAXN];
char num[MAXN][MAXN];
int n,m;
int main()
{
int i,j,k,t,bz;
int cas=0;
scanf("%d",&t);
while(t--)
{
bz=0;
scanf("%d",&n);
memset(pri,0,sizeof(pri));
memset(v,0,sizeof(v));
for(i=0;i<n;i++)
{
scanf("%s",num[i]);
for(j=0;j<n;j++)
{
if(num[i][j]=='1')
{
pri[i+1][j+1]=1;
v[j+1]++;
}
}
}
for(i=1;i<=n;i++)
{
k=0;
for(j=1;j<=n;j++)
{
if(v[j]==0)
{
k=j;
break;
}
}
if(k==0)
{
bz=1;
break;
}
else
{
v[k]--;
for(j=1;j<=n;j++)
{
if(pri[k][j]==1)
v[j]--;
}
}
}
printf("Case #%d: ",++cas);
if(bz==0)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}