HDU 5631 （BestCoder Round #73 (div.1) 1001）Rikka with Graph（并查集）

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 51    Accepted Submission(s): 20

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.

Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number T(T≤30)——The
number of the testcases.

For each testcase, the first line contains a number n(n≤100).

Then n+1 lines follow. Each line contains two numbers u,v ,
which means there is an edge between u and v.

Output

For each testcase, print a single number.

Sample Input

1
3
1 2
2 3
3 1
1 3

Sample Output

9

大致题意：

有n个点，n+1条边，最少去掉一条边使得构成的图是联通图

方法：

n个点最少是有n-1条边才能联通，所以枚举去掉一条和去掉两条的边的情况

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000")
#define LL long long
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const double eqs=1e-9;
const int MAXN=150010;

int u[110],v[110];
int f[110];

bool Judge(int n)
{

int num = 0;
for(int i = 1; i <= n; i++)
if(f[i] == i)
num++;
if(num == 1)
return true;
return false;

}

int Find(int x)
{
return x == f[x] ? x : f[x] = Find(f[x]);
}
void Init(int n)
{
for(int i=0;i<=n;i++)
f[i] = i;

}

void Link(int a, int b)
{
int fx = Find(a);
int fy = Find(b);
if(fx != fy)
f[fx] = fy;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
int ant = 0;
scanf("%d",&n);
for(int i=0; i<=n; i++)
{
scanf("%d %d",&u[i],&v[i]);
}
for(int i=0; i<n; i++)
{
for(int j=i+1; j<=n; j++)
{
Init(n);
for(int k=0; k<=n; k++)
{
if(k!=i && k!=j)
Link(u[k], v[k]);
}

if(Judge(n))
ant++;
}
}
for(int i=0;i<=n;i++)
{
Init(n);
for(int j=0;j<=n;j++)
{
if(i!=j)
{
Link(u[j], v[j]);
}

}if(Judge(n))
ant++;
}
printf("%d\n",ant);
}
return 0;
}