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): 3692    Accepted Submission(s): 1455


[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

 

 

//对于矩阵来说，第i行第j列是1时，表示i喜欢j，给你一个爱情关系矩阵，判断是否存在三角恋，或者是多角恋，

爱情啊，真让人捉摸不透

 

 

#include<cstdio>
#include<cstring>
char map[2005][2005];
int times[2005];
int t=1;
void topo(int n)
{
int i,j,top;
for (i=0;i<n;i++)
{
top=-1;
for (j=0;j<n;j++)
{
if (times[j]==0)
{
top=j;
times[j]--;
break;
}
}
if (top==-1)
break;
for (j=0;j<n;j++)
{
if (map[top][j]=='1')
{
times[j]--;
map[top][j]='0';
}
}
}
printf ("Case #%d:",t++);
if (i<n)
printf (" Yes\n");
else
printf (" No\n");
}
int main()
{
int k,m,n,i,j;
scanf ("%d",&k);
while (k--)
{
scanf ("%d",&n);
memset(times,0,sizeof(times));
for (i=0;i<n;i++)
{
scanf ("%s",map[i]);
for (j=0;j<n;j++)
{
if (map[i][j]=='1')
times[j]++;
}
}
topo(n);
}
return 0;
}

 

 

 

 

 

 

 

 

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