ZOJ Friends（暴力）

Alice lives in the country where people like to make friends. The friendship is bidirectional and if any two person have no less thank friends in common, they will become friends in several days. Currently, there are totallyn 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. ThenT 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 integersu_i, v_i (0 ≤ u_i, v_i <n, u_i ≠ v_i) indicating there is friendship between personu_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

34 4 20 10 21 32 35 5 20 11 22 33 44 05 6 20 11 22 33 44 02 0

Sample Output

204

题意: n个人，有m个朋友关系（双向），如果两个人有num个以上的（含num）个公共朋友，那么他们也是朋友，问可以新加几个朋友关系

思路: 暴力，枚举i j之间有多少个公共朋友

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8#define fre(i,a,b)  for(i = a; i <= b; i++)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 1000000000#define N 105int  n,m,num,ans;int a[N][N];void solve(){int i,j,cnt;ans=0;while(1){  bool flag=false;  for(i=0;i<n;i++)for(j=0;j<n;j++)  {  if(i==j) continue;  if(a[i][j]) continue;                int cnt=0;    for(int k=0;k<n;k++)if(a[i][k]&&a[k][j]) cnt++;if(cnt>=num){a[i][j]=a[j][i]=1;ans++;    flag=true;}  }          if(!flag)  break;}printf("%d\n",ans);}int main(){int i,j,t;ans=0;sf(t);while(t--){sfff(n,m,num);mem(a,0);int s,e;while(m--){sff(s,e);a[s][e]=a[e][s]=1;}  solve();}return 0;}