zoj3710 friends（floyd变形）

Friends

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Time Limit: 2 Seconds     
Memory Limit: 65536 KB

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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

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

大概思路：

1，数学方法？

2，二进制求交集+搜索+减枝？

3，暴力枚举？

大部分人用的floyd思想递推 ：

i认识x，j认识x，那么i就有可能和j是朋友 ，找到i认识j的路径（x数）多等于k即可，如果可以认识，我们动态建路即可

对floyd的理解

for(k=1;k<=n;k++)

for(i=1;i<=n;i++)

for(j=1;j<=n;j++)

if(...)...

其中k为最对循环次数，我们可以改成：

while(上一次有更新)

{

for(i=1;i<=n;i++)

for(j=1;j<=n;j++)

if(...)...+标记更新

}

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<memory.h>
using namespace std;
bool map[101][101];
int n,m,k,ans;
void _update()
{
     ans=0;
	 memset(map,false,sizeof(map));
}
void _in()
{
	    int x,y;
    	scanf("%d%d%d",&n,&m,&k);
    	for(int i=0;i<m;i++){
		    scanf("%d%d",&x,&y);
		    map[x][y]=map[y][x]=true;
		}
}
void _solve()
{
	bool temp=true;
	while(temp)
	{
		temp=false;
		for(int i=0;i<n;i++)
		 for(int j=i+1;j<n;j++)
		 {
				if(map[i][j]) continue;
				int cnt=0;
				for(int f=0;f<n;f++)
				if(map[i][f]&&map[j][f])
				   cnt++;
				   if(cnt>=k) {
					ans++;
					map[i][j]=map[j][i]=true;
					temp=true;
				  }
		 }
	}
    printf("%d\n",ans);
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--){
		_update();
		_in();
		_solve();
	}
}