题目大意：给你n个点然后让你求出去掉一条边之后所形成的最小生成树。

比较基础的次小生成树吧。。。先prime一遍求出最小生成树，在dfs求出次小生成树。

Install Air Conditioning

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Problem Description

NJUST carries on the tradition of HaJunGong. NJUST, who keeps up the ”people-oriented, harmonious development” of the educational philosophy and develops the ”unity, dedication, truth-seeking, innovation” school motto, has now become an engineering-based,
multidisciplinary university.

As we all know, Nanjing is one of the four hottest cities in China. Students in NJUST find it hard to fall asleep during hot summer every year. They will never, however, suffer from that hot this year, which makes them really excited. NJUST’s 60th birthday
is approaching, in the meantime, 50 million is spent to install air conditioning among students dormitories. Due to NJUST’s long history, the old circuits are not capable to carry heavy load, so it is necessary to set new high-load wires. To reduce cost, every
wire between two dormitory is considered a segment. Now, known about all the location of dormitories and a power plant, and the cost of high-load wire per meter, Tom200 wants to know in advance, under the premise of all dormitories being able to supply electricity,
the minimum cost be spent on high-load wires. And this is the minimum strategy. But Tom200 is informed that there are so many wires between two specific dormitories that we cannot set a new high-load wire between these two, otherwise it may have potential
risks. The problem is that Tom200 doesn’t know exactly which two dormitories until the setting process is started. So according to the minimum strategy described above, how much cost at most you‘ll spend?

Input

The first line of the input contains a single integer T(T ≤ 100), the number of test cases.

For each case, the first line contains two integers n(3 ≤ n ≤ 1000), k(1 ≤ k ≤ 100). n represents n-1 dormitories and one power plant, k represents the cost of high-load wire per meter. n lines followed contains two integers x, y(0 ≤ x, y ≤ 10000000), representing
the location of dormitory or power plant. Assume no two locations are the same, and no three locations are on a straight line. The first one is always the location of the power plant.

Output

For each case, output the cost, correct to two decimal places.

Sample Input

2

4 2
0 0
1 1
2 0
3 1

4 3
0 0
1 1
1 0
0 1

Sample Output

9.66
9.00

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

#define LL __int64
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 1010;
const int N = 1010;
const int M = 300010;
struct node
{
    int x, y;
} p[maxn];

struct node1
{
    int to, next;
} f[M];

int pre[N], head[N], flag[N];
int vis[N][N];
double low[N];
double dis[N][N], dp[N][N];
int n, m, num;
double sum;

void init()
{
    sum = 0;
    num = 0;
    memset(flag, 0, sizeof(flag));
    memset(vis, 0, sizeof(vis));
    memset(head, -1, sizeof(head));
    memset(dp, 0, sizeof(dp));
}

double Dis(int x0, int y0, int x1, int y1)
{
    return sqrt(1.0*(x0-x1)*(x0-x1) + 1.0*(y0-y1)*(y0-y1));
}

void add(int s, int t)
{
    f[num].to = t;
    f[num].next = head[s];
    head[s] = num ++;
}

void prime()
{
    for(int i = 1; i <= n; i++)
    {
        low[i] = dis[1][i];
        pre[i] = 1;
    }
    flag[1] = 1;
    for(int i = 1; i < n; i++)
    {
        double Min = INF;
        int v;
        for(int j = 1; j <= n; j++)
        {
            if(!flag[j] && Min > low[j])
            {
                v = j;
                Min = low[j];
            }
        }
        sum += Min;
        vis[pre[v]][v] = vis[v][pre[v]] = 1;
        add(v, pre[v]);
        add(pre[v], v);
        flag[v] = 1;
        for(int j = 1; j <= n; j++)
        {
            if(!flag[j] && low[j] > dis[v][j])
            {
                low[j] = dis[v][j];
                pre[j] = v;
            }
        }
    }
}

double dfs(int cur, int u, int fa)  //用cur更新cur点所在的子树和另外子树的最短距离
{
    double ans = INF;
    for(int i = head[u]; ~i; i = f[i].next) //沿着生成树的边遍历
    {
        if(f[i].to == fa)
            continue;
        double tmp = dfs(cur, f[i].to, u);  //用cur更新的以当前边为割边的两个子树最短距离
        ans = min(tmp, ans);        //以(fa,u)为割边的2个子树的最短距离
        dp[u][f[i].to] = dp[f[i].to][u] = min(tmp, dp[u][f[i].to]);
    }
    if(cur != fa)   //生成树边不更新
        ans = min(ans, dis[cur][u]);
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d %d",&n, &m);
        for(int i = 1; i <= n; i++) scanf("%d %d",&p[i].x, &p[i].y);
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= i; j++)
            {
                dp[i][j] = dp[j][i] = INF;
                if(i == j)
                {
                    dis[i][j] = 0.0;
                    continue;
                }
                dis[i][j] = dis[j][i] = Dis(p[i].x, p[i].y, p[j].x, p[j].y);
            }
        }
        prime();
        double ans = sum;
        for(int i = 0; i < n; i++) dfs(i, i, -1);
        for(int i = 2; i <= n; i++)
        {
            for(int j = 2; j < i; j++)
            {
                if(!vis[i][j]) continue;
                ans = max(ans, sum-dis[i][j]+dp[i][j]);
            }
        }
        printf("%.2lf\n",ans*m);
    }
    return 0;
}