HDU 4756 次小生成树裸题

最新推荐文章于 2021-04-09 21:35:56 发布

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最新推荐文章于 2021-04-09 21:35:56 发布

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分类专栏：基础图论

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南京大学为解决学生宿舍夏季高温问题，投入巨资安装空调。文章探讨了在确保所有宿舍都能供电的前提下，如何在旧电路负荷限制下，通过设定新高负载电线来最小化成本，并详细介绍了次小生成树算法在该问题中的应用。

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Install Air Conditioning

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 768 Accepted Submission(s): 154

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

Sample Output

9.66
9.00


     南京网赛时的题目，看出来是次小生成树，不过数据规模是1000，显然只能n^2的做法，自己不会，网上也没搜到，只好没能做。
     次小生成树的解法：
     首先对原图求一次最小生成树，把最小生成树上的边标记出来，然后树形dp枚举去掉最小生成树上的每一条边后的最小生成树，然后扫一遍就是答案。
代码：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <climits>
#include <cmath>
#include <cassert>
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <deque>
#include <algorithm>
using namespace std;
typedef long long ll;
const double eps=1e-8;
const int maxn=1010;
struct point
{
       double x,y;
       void input()
       {
              scanf("%lf%lf",&x,&y);
       }
}pp[maxn];
double dist(point a,point b)
{
       double sss=a.x-b.x;
       double ttt=a.y-b.y;
       return sqrt(sss*sss+ttt*ttt);
}
const double inf=1000000000;
double dis[maxn][maxn],dp[maxn][maxn],lowc[maxn];
int head[maxn],tol,vis[maxn],mp[maxn][maxn],pre[maxn];
struct node
{
       int next,to;
}edge[maxn*maxn];
void add(int u,int v)
{
       edge[tol].to=v;
       edge[tol].next=head[u];
       head[u]=tol++;
}
double prim(int n)
{
       memset(vis,0,sizeof(vis));
       memset(mp,0,sizeof(mp));
       memset(pre,0,sizeof(pre));
       double ans=0;
       vis[0]=1;pre[0]=-1;lowc[0]=0;
       for(int i=1;i<n;i++)lowc[i]=dis[0][i];
       for(int i=1;i<n;i++)
       {
              double minc=inf;
              int p=-1;
              for(int j=0;j<n;j++)
              if(!vis[j]&&minc>lowc[j])
              {
                     minc=lowc[j];
                     p=j;
              }
              ans+=minc;
              vis[p]=1;
              mp[pre[p]][p]=mp[p][pre[p]]=1;
              add(pre[p],p);add(p,pre[p]);
              for(int j=0;j<n;j++)
              if(!vis[j]&&lowc[j]>dis[p][j])
              {
                     lowc[j]=dis[p][j];
                     pre[j]=p;
              }
       }
       return ans;
}
double dfs(int cur,int u,int fa)
{
       double res=inf;
       for(int i=head[u];i!=-1;i=edge[i].next)
       {
              int v=edge[i].to;
              if(v==fa)continue;
              double tmp=dfs(cur,v,u);
              dp[u][v]=dp[v][u]=min(dp[u][v],tmp);
              res=min(res,tmp);
       }
       if(fa!=cur)res=min(res,dis[cur][u]);
       return res;
}
int main()
{
       int T,i,j,k,m,n;
       scanf("%d",&T);
       while(T--)
       {
              scanf("%d%d",&n,&k);
              memset(head,-1,sizeof(head));tol=0;
              for(i=0;i<n;i++)pp[i].input();
              for(i=0;i<n;i++)
              {
                     dis[i][i]=0;
                     for(j=i+1;j<n;j++)
                            dis[i][j]=dis[j][i]=dist(pp[i],pp[j]);
              }
              double ANS=prim(n);
              //cout<<ANS<<endl;
              for(i=0;i<n;i++)
                     for(j=0;j<n;j++)
                            dp[i][j]=inf;
                     for(i=0;i<n;i++)dfs(i,i,-1);
                     //cout<<tol<<endl;
              double ans=ANS;
              for(i=1;i<n;i++)
                     for(j=i+1;j<n;j++)
                     if(mp[i][j])
                     ans=max(ans,ANS-dis[i][j]+dp[i][j]);
                     printf("%.2lf\n",k*ans);
       }
       return 0;
}