zoj 3710 Friends（数学）

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本文详细解读了Zhejiang省高校编程竞赛中的一个算法问题，即如何通过计算朋友关系来判断新友谊的形成。文章通过提供输入输出示例，阐述了解决此类问题的思路和算法实现，包括如何利用邻接矩阵来表示人际关系，以及如何遍历矩阵以找出满足特定条件的新友谊。此外，还提供了代码示例，展示了如何将理论转化为实际的编程解决方案。

转载请注明出处：http://blog.youkuaiyun.com/u012860063?viewmode=contents

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3710

Friends

Time Limit: 2 Seconds Memory Limit: 65536 KB

Alice lives in the country where people like to make friends. The friendship is bidirectional and if any two person have no less than k friends in common, they will become friends in several days. Currently, there are totally n people in the country, and m friendship among them. Assume that any new friendship is made only when they have sufficient friends in common mentioned above, you are to tell how many new friendship are made after a sufficiently long time.

Input

There are multiple test cases.

The first lien of the input contains an integer T (about 100) indicating the number of test cases. Then T cases follow. For each case, the first line contains three integersn, m, k (1 ≤ n ≤ 100, 0 ≤ m ≤ n×(n-1)/2, 0 ≤ k ≤ n, there will be no duplicated friendship) followed by m lines showing the current friendship. The ith friendship contains two integers u_i, v_i (0 ≤ u_i, v_i < n, u_i ≠ v_i) indicating there is friendship between person u_i and v_i.

Note: The edges in test data are generated randomly.

Output

For each case, print one line containing the answer.

Sample Input

Sample Output

2
0
4

Author: ZHUANG, Junyuan

Contest: The 10th Zhejiang Provincial Collegiate Programming Contest

题意：

两个人有大于等于K个相同的朋友!那么他们也算是朋友！问一共有多少对这样的新友谊！

PS：

一旦有新友谊后!那么计算后面的是否有大于等于K对朋友的时候，新友谊也可以考虑在其中！

代码如下：

#include<cstdio>
#include<cstring>
int main()
{
	int t,n,m,k,x,i,j,l,R,L,num,s;
	int map[147][147];
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			num=0;
			scanf("%d%d%d",&n,&m,&k);
			memset(map,0,sizeof(map));
			for(i=0;i<m;i++)
			{
				scanf("%d%d",&R,&L);
				map[R][L]=1;
				map[L][R]=1;
			}
			int sign = 1 ;
			while( sign)
			{
				sign = 0 ;
				for( i = 0 ; i < n ; i++ )
				{
					for( j = i+1 ; j < n ; j++ )
					{
						s = 0;
						if(map[i][j] == 1)//已经是朋友的不再计算
							continue;
						for(l = 0 ; l < n ; l++)
						{
							if(map[i][l] == 1 && map[j][l] == 1)
								s++;
						}
						if(s >= k)
						{
							num++;
							map[i][j] = 1;//此处标记新友谊，以便后面使用
							map[j][i] = 1;
							sign  = 1;
						}
					}
				}
			}
			printf("%d\n",num);
		}
	}
	return 0;
}