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ZOJ - 3710 Friends 暴力
阅读量:3904 次
发布时间:2019-05-23

本文共 2001 字,大约阅读时间需要 6 分钟。

题目:

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 ui, vi (0 ≤ ui, vi < n, ui ≠ vi) indicating there is friendship between person ui and vi.

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

思路:

暴力,直到没有新的朋友关系出现为止。

代码如下:

#include 
#include
#include
#include
using namespace std;const int maxn=1e2+5;int t;int n,m,k;int a[maxn][maxn];int ans;int main(){ scanf("%d",&t); while(t--) { ans=0; memset (a,0,sizeof(a)); scanf("%d%d%d",&n,&m,&k); for (int i=1;i<=m;i++) { int x,y; scanf("%d%d",&x,&y); a[x][y]=a[y][x]=1; } while(1) { int flag=0; for (int i=0;i
=k) { flag=1; ans++; a[j][i]=a[i][j]=1; } } } } if(flag==0) break; } printf("%d\n",ans); } return 0;}

 

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