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
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
题解：题目大意，T组数据，N个朋友，M个友谊关系，关键值K，接下来M行A，B。表示A，b是好友，任意两个人之间只要共同好友数大于等于K，则他们成为新朋友，求，经过一段时间后，新朋友的数目……
我一开始用的结构体，但忘了用while，结果，一直错，加上while之后，过了，但看了看别人的代码，发现我写的好复杂……
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<string>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define ll long long
#define For(i,a,b) for(int i=a;i<b;i++)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
struct node
{
int a,b[105]; // b 存与x[i]是朋友关系的人，a 存b 的大小
} x[105];
int c[205];
int v[105][105]; // 做下标记
int main()
{
int t,n,m,M;
scanf("%d",&t);
while(t--)
{
mem(v,0);
scanf("%d%d%d",&n,&m,&M);
for(int i=0; i<n; i++)
x[i].a=0;
int aa,bb;
For(i,0,m)
{
scanf("%d%d",&aa,&bb);
x[aa].b[x[aa].a++]=bb;
x[bb].b[x[bb].a++]=aa;
v[aa][bb]=v[bb][aa]=1;
}
int ans=0,falg;
while(1) // 防止新朋友出现后，依据新朋友又出现了新朋友
{
falg=0;
for(int i=0; i<n; i++)
{
if(x[i].a<M)
continue;
for(int j=0; j<n; j++)
{
if(j==i)
continue;
if(v[i][j]||v[j][i])
continue;
if(x[j].a<M)
continue;
mem(c,0);
sort(x[i].b,x[i].b+x[i].a);
int aaa=unique(x[i].b,x[i].b+x[i].a)-x[i].b;
for(int k=0; k<aaa; k++)
c[k]=x[i].b[k];
sort(x[j].b,x[j].b+x[j].a);
int bbb=unique(x[j].b,x[j].b+x[j].a)-x[j].b;
for(int k=aaa; k<aaa+bbb; k++)
c[k]=x[j].b[k-aaa];
sort(c,c+aaa+bbb);
int d=unique(c,c+aaa+bbb)-c;
if(aaa+bbb-d>=M)
{
x[i].b[x[i].a++]=j;
x[j].b[x[j].a++]=i;
v[i][j]=v[j][i]=1;
ans++;
falg=1;
}
}
}
if(!falg)break;
}
printf("%d\n",ans);
}
}下面是传说中的人家的代码……
#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 105
int 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;
}还是做得题太少了，思路太僵化了，哎，努力吧……