Triangle LOVE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2138 Accepted Submission(s): 898
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). 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).
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
2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110
Sample Output
Author
BJTU
题解:
本题要求给定的有向图里是否又a->b->c->a的三角形环。由于矩阵中任意两点有么有正向边,要么有反向边,所以只要有环,就能构成三角形环。证明如下:假设有n元环,a->b->c……->a,若c->a有边,则存在三角形环a->b->c->a;若无边,则说明不存在c->a的边从而存在a->c的边,于是有n元环a->b->c……->a变成有n-1元环,a->c……->a,最终可以证明存在三角形环。时间复杂度为拓扑排序的O(n*n),在本题可以接受。
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN=2000+10;
int inDev[MAXN];
bool visited[MAXN];
bool g[MAXN][MAXN];
char str[MAXN];
int n;
void init()
{
memset(g,0,sizeof(g));
memset(visited,0,sizeof(visited));
memset(inDev,0,sizeof(inDev));
}
void input()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%s",str);
for(int j=0;j<n;j++)
{
if('1'==str[j])
{
g[i][j]=true;
++inDev[j];
}
}
}
}
void topSort()
{
int i,tag=0,k=0;
while(k<n)
{
for(i=0;i<n;i++)
if(!visited[i]&&0==inDev[i])
break;
if(i>=n)
{
tag=1;
break;
}
++k;
visited[i]=true;
for(int j=0;j<n;j++)
if(g[i][j]&&inDev[j]!=0)
--inDev[j];
}
if(tag)
printf("Yes\n");
else printf("No\n");
}
int main()
{
int cas,tag=0;
cin>>cas;
while(cas--)
{
init();
input();
printf("Case #%d: ",++tag);
topSort();
}
system("pause");
return 0;
}
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hdu4324(拓扑排序)
原文:http://blog.csdn.net/xiaofengcanyuexj/article/details/29179639