hdoj 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3903    Accepted Submission(s): 1537

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
注意。注意，注意，取边，用字符串，不然超时！！！
#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;#include<string.h>#include<queue>#define N  2010struct line{    int u,v,next;}edge[N*N];int n,m,inde[N],head[N],top=0;void add(int u,int v){    edge[top].u =u;    edge[top].v =v;    edge[top].next =head[u];    head[u]=top++;    inde[v]++;}void topo(){    int i,j,t=0,k=0,s;    queue<int>Q;    for(i=0;i<n;i++)     {         if(inde[i]==0)//入度为零入队          {             Q.push(i);         }     }     while(!Q.empty() )     {         t=Q.front() ;         Q.pop() ;         inde[t]=-1;         k++;//记录入队人数          for(i=head[t];i!=-1;i=edge[i].next )         {              s=edge[i].v ;             inde[s]--;             if(inde[s]==0)             Q.push(s);         }     }     if(k<n)//判断是否有环      printf("Yes\n");     else     printf("No\n");}int main(){      int i,j,u,v=1;    scanf("%d",&m);    while(m--)    {        top=0;        memset(inde,0,sizeof(inde));        memset(head,-1,sizeof(head));        scanf("%d",&n);         getchar();         char a[N];         for(i=0;i<n;i++)//必须用字符串，直接取数字超时           {              scanf("%s",a);              for(j=0;j<n;j++)               {                   if(a[j]=='1')                   add(i,j);               }          }          printf("Case #%d: ",v++);            topo();                }    return 0;}