第二次周赛 friend

Problem Description

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

3
4 4 2
0 1
0 2
1 3
2 3
5 5 2
0 1
1 2
2 3
3 4
4 0
5 6 2
0 1
1 2
2 3
3 4
4 0
2 0

Sample Output

2
0
4

#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#define INF 10000000
using namespace std;
vector <int > g[102];


int vis[102],d[102],p[102],w[102][102],rank[102];//w[a][b]  a到b的 优惠（为  - ）
void spfa(int s,int n)
{
    int i;
    for(i=1;i<=n;i++)
        d[i]=INF;
    d[s]=10000;
    memset(vis,0,sizeof(vis));
    queue<int > q;
    q.push (s); vis[s]=1;
    while(!q.empty ())
    {
        int a,b;
        a=q.front (); q.pop ();
        for(i=0;i<g[a].size ();i++)
        {
            b=g[a][i];
            if(d[b]>d[a]+w[a][b])
            {
                d[b]=d[a]+w[a][b];
                if(!vis[b])
                {
                    vis[b]=1;
                    q.push (b);
                }
            }
        }
    }
}


int main()
{
    int i,j,m,n,p[102],rank[102],x,a,b;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(w,0,sizeof(w));   //    一定会有优惠  即w一定会《0
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&p[i],&rank[i],&x);
            while(x--)
            {
                scanf("%d%d",&a,&b);
                w[i][a]=b-p[i];      //a  i优惠的金币  为-
            }
        }
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                if(abs(rank[i]-rank[j])<=m&&w[i][j]!=0)
                {
                    w[i][j]+=p[j];
                    g[i].push_back (j);
                }
            }
            /*for(i=1;i<=n;i++)
            {
                printf("%d    ",i);
                for(j=0;j<g[a].size ();j++)
                    printf("%d ",g[a][j]);
                printf("
");
            }*/


         spfa(1,n);
         //for(i=1;i<=n;i++)
        //     printf("%d  d %d
",i,d[i]);
         int minn=INF;
         for(i=1;i<=n;i++)
             if(minn>d[i])
                 minn=d[i];
             if(minn<0)
                 minn=0;
        printf("%d
",minn);
    }
    return 0;
}