題目大意：給你n個點然後讓你求出去掉一條邊之後所形成的最小生成樹。

比較基礎的次小生成樹吧。。。先prime一遍求出最小生成樹，在dfs求出次小生成樹。

Install Air Conditioning

  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?

#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點所在的子樹和另外子樹的最短距離
{
    double ans = INF;
    for(int i = head[u]; ~i; i = f[i].next) //沿著生成樹的邊遍歷
    {
        if(f[i].to == fa)
            continue;
        double tmp = dfs(cur, f[i].to, u);  //用cur更新的以當前邊為割邊的兩個子樹最短距離
        ans = min(tmp, ans);        //以(fa,u)為割邊的2個子樹的最短距離
        dp[u][f[i].to] = dp[f[i].to][u] = min(tmp, dp[u][f[i].to]);
    }
    if(cur != fa)   //生成樹邊不更新
        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;
}