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hdu 4324 Triangle LOVE

2024-07-14 13:10:17   219人阅读

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2306    Accepted Submission(s): 960


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
25001001000001001111011100050111100000010000110001110
 

 

Sample Output
Case #1: YesCase #2: No
 

 

 

 

            题意：给你一个关系矩阵，比如i行j列的数值为1，带表i喜欢j。如果i喜欢j，则j一定不喜欢i，而且任何2个人之间都存在关系，要你判断是否存在一个三元环，A喜欢B，B喜欢C，C喜欢A。

           这道题我WA+MLE+TLE,最终还是没有过吧。本来想用DFS的，可惜超时了。

            

          网上的解题思路：此题可以一遍拓扑排序判环求解 即只需要找到一个环，就必定存在三元环 证明如下： 假设存在一个n元环，因为a->b有边，b->a必定没边，反之也成立所以假设有环上三个相邻的点a-> b-> c,那么如果c->a间有边，就已经形成了一个三元环，如果c->a没边，那么a->c肯定有边，这样就形成了一个n-1元环。。。。所以只需证明n大于3时一定有三元环即可，显然成立。

 

代码：

#include <stdio.h>#include <string.h>#define MAXN 2005int Indegree[MAXN];char str[MAXN][MAXN];int main(){    int n, x, num, t, T, mark;    scanf("%d", &T);    for(int k = 1; k<=T; k++)    {        memset(Indegree, 0, sizeof(Indegree));        mark = 0;        scanf("%d", &n);        for(int i = 0; i<n; i++)        {            scanf("%s", str[i]);            for(int j = 0; j<n; j++)            {                x = str[i][j]-‘0‘;                if(x)                {                    Indegree[j]++;  //记录每个节点的入度                }            }        }        for(int i = 0; i<n; i++)        {            for(t = 0; t<n; t++)   //若入度为0，则切断与它相连的所以路径            {                if(Indegree[t] == 0)                    break;            }            if(t == n)     //形成了环。            {                mark = 1;                break ;            }            else            {                for(int j = 0; j<n; j++)  //切断操作。                {                    if(str[t][j] == ‘1‘)                    {                        Indegree[j]--;                    }                }            }        }        if(mark)            printf("Case #%d: Yes\n", k);        else            printf("Case #%d: No\n", k);    }    return 0;}

 


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