hdu 4085 Peach Blossom Spring（斯坦纳树）_the peach blossom spring reflected the yearning of-优快云博客

本文链接：https://blog.youkuaiyun.com/clover_hxy/article/details/51526396

Peach Blossom Spring

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2368 Accepted Submission(s): 906

Problem Description

Tao Yuanming(365-427) was a Chinese poet of Eastern Jin dynasty. One of his most famous works is "Peach Blossom Spring", which is a fable about a chance
discovery of an ethereal village where the people lead an ideal existence in harmony with nature, unaware of the outside world for centuries. So in Chinese, "Peach Blossom Spring" means "utopia".
In the story of "Peach Blossom Spring", there was a mysterious place. In Qin dynasty, some people escaped to that place during the civil unrest and built a village. They and their descendants never left and never had any contact with the outside world since then, until centuries latter a fisherman of Jin dynasty found them.
Recently, some Chinese ACMers happened to find the relics of the village mentioned in"Peach Blossom Spring".
They also found a document about building hiding places to escape from Qin army. The document said:
There were n houses and m roads in the village. Each road connected two houses. These houses were numbered from 1 to n. There were k families, each living in a different house.
The houses they lived were house 1, house 2, … , house k. There were also k broken houses: house n-k+1, house n-k+2, ... , house n, with secret basements so that those houses could be used as hiding places.
The problem was that all roads were broken. People wanted to repair some roads so that every family could reach a hiding place through the repaired roads. Every hiding place could only hold one family. Each road cost some labor to be repaired. The head of the village wanted to find out the minimum cost way of repairing the roads, but he didn't know how to do.
Would you solve the problem which the ancient village head never solved?

Input

The input begins with a line containing an integer T(T<=50), the number of test cases. For each case, the first line begins with three integers ---- the above mentioned n (4<=n<=50), m (0<=m<=1000) and k (1<=k<=5, 2k<=n). Then m lines follow, each containing three integers u,v and w, indicating that there is a broken road connecting house u an d v, and the cost to repair that road is w(1<=w<=1000).

Output

For each test case, if you cannot find a proper way to repair the roads, output a string "No solution" in a line. Otherwise, output the minimum cost to repair the roads in a line.

Sample Input

Sample Output

Source

2011 Asia Beijing Regional Contest

Recommend

lcy | We have carefully selected several similar problems for you: 4089 4081 4082 4090 4087

Statistic | Submit | Discuss | Note

题目大意：给你n,m,k ，分别表示有n个点，m条边，每条边有一个权值，表示修复这条边需要的代价，从前k个点中任取一个使其和后k个点中的某一个点，通过边连接，并且必须是一一对应，问最小的代价是多少。

题解：刚开始的时候看到某些点必须选并使其联通，所以想到了斯坦纳树，但是需要注意的是这道题是让k 对点满足一一对应的联通关系，也就是说最后生成的不一定是一颗，而有可能是一片森林。那么我们可以先按照斯坦纳树的方式来进行DP，求出所有的状态，然后再进行一次状压dp，dp[sta]=min(dp[sta],dp[s]+dp[sta-s])，用两棵树表示当前状态。

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>
#include<queue>  
#include<cmath>  
#define N 6003  
#define inf 1000000000  
using namespace std;  
int n,m,k,tot;  
int point[N],next1[N],v[N],len[N];  
int f[53][(1<<11)],mi[20],can[N],dp[(1<<11)];  
queue<int> p;
void add(int x,int y,int z)  
{  
    tot++; next1[tot]=point[x]; point[x]=tot; v[tot]=y; len[tot]=z;  
    tot++; next1[tot]=point[y]; point[y]=tot; v[tot]=x; len[tot]=z;  
}  
void spfa(int sta)
{
	while(!p.empty())
	{
		int now=p.front(); p.pop();
		for (int i=point[now];i;i=next1[i])
		{
			if (f[v[i]][sta]>f[now][sta]+len[i])
			{
				f[v[i]][sta]=f[now][sta]+len[i];
				if (!can[v[i]]){
					can[v[i]]=1;
					p.push(v[i]);
				}
			}
		}
		can[now]=0;
	}
}
bool check(int sta)//判断当前的状态是否满足一一对应关系
{
	int ans=0;
	for (int i=0;i<k;i++)
	{
	  if (sta&(1<<i))  ans++;
	  if (sta&(1<<(i+k))) ans--;
	}
	return (ans==0);
}
int main()  
{  
    int t; scanf("%d",&t);  
    mi[0]=1;  
    for (int i=1;i<=11;i++)  mi[i]=mi[i-1]*2; 
    for (int T=1;T<=t;T++)  
    {  
        scanf("%d%d%d",&n,&m,&k);  
        tot=0; 
        memset(point,0,sizeof(point));
        memset(next1,0,sizeof(next1));
		for (int i=1;i<=m;i++)  
        {  
            int x,y,z;  scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);  
        }  
        for (int i=1;i<=n;i++)  
         for (int j=0;j<mi[10];j++)
		   f[i][j]=inf;
		for (int i=1;i<=k;i++)
		  f[i][mi[i-1]]=0;
		int t=k;
		for (int i=n-k+1;i<=n;i++)
		 f[i][mi[t]]=0,t++;
		for (int sta=1;sta<mi[t];sta++)
		{
			for (int i=1;i<=n;i++)
			{
				for (int s=sta&(sta-1);s;s=sta&(s-1))
				{
					int t=f[i][sta-s]+f[i][s];
					f[i][sta]=min(f[i][sta],t);
				}
				if (f[i][sta]!=inf)  p.push(i),can[i]=1;
			}
			spfa(sta);
		}  
		for (int sta=1;sta<mi[t];sta++)
		{
			dp[sta]=inf;
			for (int i=1;i<=n;i++)
			 dp[sta]=min(dp[sta],f[i][sta]);
		}
		for (int sta=1;sta<mi[t];sta++)
		 if (check(sta))
		  for (int s=sta&(sta-1);s;s=sta&(s-1))
		   if (check(s))  dp[sta]=min(dp[sta],dp[s]+dp[sta-s]);
		if (dp[mi[t]-1]==inf)  printf("No solution\n");
		else   printf("%d\n",dp[mi[t]-1]);
    }  
}