Friends ZOJ - 3710（交集）

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

用set存边
然后如果两个点的交集大于等于k
则说明这两个可以连边
再用set存一下
重新开始判断（为什么  自己想想）

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 110000, INF = 0x7fffffff;

int n, m, k;
set<int> s[105];
set<int> ss;
int vis[105][105];
int main()
{
    int T;
    rd(T);
    while(T--)
    {
        int u, v;
        rd(n), rd(m), rd(k);
        for(int i = 0; i <= n; i++)
        {
            s[i].clear();
            for(int j = 0; j <= n; j++) vis[i][j] = 0;
        }
        for(int i = 0; i < m; i++)
        {
            rd(u), rd(v);
            s[u].insert(v), s[v].insert(u);
            vis[u][v] = vis[v][u] = 1;
        }
        int cnt = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(i == j) continue;
                if(vis[i][j]) continue;
                if(s[i].size() < k || s[j].size() < k) continue;
                ss.clear();
                set_intersection(s[i].begin(), s[i].end(), s[j].begin(), s[j].end(), inserter(ss, ss.begin()));
                if(ss.size() >= k)
                {
                    cnt++;
                    vis[i][j] = vis[j][i] = 1;
                    s[i].insert(j);
                    s[j].insert(i);
                    i = j = 0;
                }
            }
        }
        pd(cnt);

    }


    return 0;
}

标签：3710,int,friendship,ZOJ,rd,include,Friends,scanf,define
来源： https://www.cnblogs.com/WTSRUVF/p/10802643.html