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hdu 4324

2024-07-16 05:02:58   213人阅读

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2478    Accepted Submission(s): 1011


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
Case #1: Yes
Case #2: No

 对于两个任何两个点都会有一条边将他们相连（Ai,j ≠ Aj,i）

假设目前有一个环大于3，其中有相连的三个点A,B,C，即 A->B,B->C

如果C->A 有一条边，就形成了一个3环

如果C->A 没有边，那么一定有 A->C，从而形成一个n-1环，以这种方式一定可以形成一个3环

所以只要判断这些点构成一个环即可

 1 #include<string> 2 #include<cstdio> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 #include<cstring> 9 #include<stdlib.h>10 #include<string>11 #include<cmath>12 using namespace std;13 #define pb push_back14 vector<int >p[10010];15 int in[10010],n,m,deep[10010],num[10010];16 char dp[2010][2010];17 void init(){18     memset(in,0,sizeof(in));19     memset(deep,0,sizeof(deep));20     memset(num,0,sizeof(num));21     for(int i=0;i<=n;i++) p[i].clear();22 }23 int tuopu(){24     queue<int >ak_47;25     int cnt=0;26     for(int i=1;i<n;i++)27     if(in[i]==0) ak_47.push(i),deep[i]=1,num[1]++;28     while(!ak_47.empty()){29         int pos=ak_47.front();30         cnt++;31         for(int i=0;i<p[pos].size();i++){32             int to=p[pos][i];33             if(--in[to]==0) ak_47.push(to),deep[to]=deep[pos]+1,num[deep[to] ]++;34         }35         ak_47.pop();36     }37     if(cnt<n) return 1;38     else return 0;39 40 }41 int main(){42     int cas=0,t;43     cin>>t;44     while(t--){45         cin>>n;46         init();47         for(int i=1;i<=n;i++){48             scanf("%s",dp[i]);49             for(int j=0;j<n;j++)50             if(dp[i][j]==‘1‘) in[j+1]++,p[i].pb(j+1);51         }52         if(tuopu()) printf("Case #%d: Yes\n",++cas);53         else   printf("Case #%d: No\n",++cas);54     }55 }

 

 


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