ZOJ 3710 Friends【floyd思想递推】

最新推荐文章于 2019-03-18 20:48:45 发布

原创最新推荐文章于 2019-03-18 20:48:45 发布 · 815 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ZOJ 3710

最短路及其拓展专栏收录该内容

166 篇文章

订阅专栏

本文探讨了一个社交网络模型，在该模型中，当两个人拥有至少k个共同朋友时，他们就会成为朋友。文章提供了一种算法来计算在一个给定的社交网络中，经过足够长的时间后能形成多少新的友谊。

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 integers n, 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

思路：如i认识j，j认识k，那么i就有可能认识 k，那么i想认识k要有什么样的条件呢？找到i认识k的路径数大等于k即可，如果可以认识，我们动态建路即可。

AC代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int map[1005][1005];
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            memset(map,0,sizeof(map));
            int n,m,k;
            scanf("%d%d%d",&n,&m,&k);
            for(int i=0; i<m; i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                map[x][y]=1;
                map[y][x]=1;
            }
            int output=0;
            int ok=1;
            while(ok)
            {
                ok=0;
                for(int i=0; i<n; i++)
                {
                    for(int j=i+1; j<n; j++)
                    {
                        if(map[i][j]==1)continue;
                        int cont=0;
                        for(int l=0; l<n; l++)
                        {
                            if(map[i][l]==1&&map[l][j]==1)
                            {
                                cont++;
                            }
                        }
                        if(cont>=k)
                        {
                            output++;
                            map[i][j]=1;
                            map[j][i]=1;
                            ok=1;
                        }
                    }
                }
            }
            printf("%d\n",output);
        }
    }
}