hdu4324

http://acm.hdu.edu.cn/showproblem.php?pid=4324

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

/**hdu4324  dfs题目大意：         判断是否存在点数为3的环。解题思路:         dfs搜索即可，不做标记会TLE，dfs时做标记如果当前点u的下一个点v已经访问过了，         那么就判断dist[u]==dist[[v]+2,成立返回true,否则更新dist[v]=dist[u]+1,继续深搜。*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;char s[2005][2005];int n,dist[2005];struct note{    int v,next;} edge[2005*2005];int head[2005],ip,flag[2005];void init(){    memset(head,-1,sizeof(head));    ip=0;}void addedge(int u,int v){    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;}bool dfs(int u){    flag[u]=1;    for(int i=head[u]; i!=-1; i=edge[i].next)    {        int v=edge[i].v;        if(flag[v]&&dist[u]==dist[v]+2)        {            puts("Yes");            return 1;        }        else if(flag[v]==0)        {            dist[v]=dist[u]+1;            if(dfs(v))                return 1;        }    }    return 0;}int main(){    int T,tt=0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=0; i<n; i++)            scanf("%s",s[i]);        init();        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                if(s[i][j]=='1')                {                    addedge(i+1,j+1);                }            }        }        memset(flag,0,sizeof(flag));        memset(dist,0,sizeof(dist));        int flag1=0;        printf("Case #%d: ",++tt);        for(int i=1; i<=n; i++)        {            if(dfs(i)==1)            {                flag1=1;                break;            }        }        if(flag1==0)            puts("No");    }    return 0;}