HDU 4085 Steiner樹模板稱號

Dig The Wells

Time Limit: 6000/2000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 971????Accepted Submission(s): 416

Problem Description

You may all know the famous story “Three monks”. Recently they find some places around their temples can been used to dig some wells. It will help them save a lot of time. But to dig the well or build the road to transport the water will cost money. They do not want to cost too much money. Now they want you to find a cheapest plan.

Input

There are several test cases.
Each test case will starts with three numbers n , m, and p in one line, n stands for the number of monks and m stands for the number of places that can been used, p stands for the number of roads between these places. The places the monks stay is signed from 1 to n then the other m places are signed as n + 1 to n + m. (1 <= n <= 5, 0 <= m <= 1000, 0 <=p <= 5000)
Then n + m numbers followed which stands for the value of digging a well in the ith place.
Then p lines followed. Each line will contains three numbers a, b, and c. means build a road between a and b will cost c.

Output

For each case, output the minimum result you can get in one line.

Sample Input

3 1 3 1 2 3 4 1 4 2 2 4 2 3 4 4 4 1 4 5 5 5 5 1 1 5 1 2 5 1 3 5 1 4 5 1

Sample Output

6 5

題意：有n個和尚。每個和尚一個廟，有m個村莊，p條路。每條路有費用，每個地方打井也須要費用，求最少花多少錢能夠使得全部和尚喝上水。

斯坦納樹比較裸的問題。

代碼：

/* ***********************************************
Author :rabbit
Created Time :2014/7/17 0:59:57
File Name :13.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 100000000
#define eps 1e-8
#define pi acosi
typedef long long ll;
int head[1100],tol;
struct Edge{int next,to,val;
}edge[1001000];
void addedge(int u,int v,int w){edge[tol].to=v;edge[tol].next=head[u];edge[tol].val=w;head[u]=tol++;
}
int weight[1100],d[1100][1<<5],dp[1100],in[1010][1<<5];
int main()
{//freopen("data.in","r",stdin);//freopen("data.out","w",stdout);int n,m,p;while(~scanf("%d%d%d",&n,&m,&p)){memset(head,-1,sizeof(head));tol=0;for(int i=0;i<n+m;i++)scanf("%d",&weight[i]);while(p--){int u,v,w;scanf("%d%d%d",&u,&v,&w);u--;v--;addedge(u,v,w);addedge(v,u,w);}for(int i=0;i<n+m;i++)for(int j=0;j<(1<<n);j++)d[i][j]=INF;for(int i=0;i<(1<<n);i++)dp[i]=INF;memset(in,0,sizeof(in));for(int i=0;i<n;i++)d[i][1<<i]=weight[i];for(int i=1;i<(1<<n);i++){queue<int> Q;for(int j=0;j<n+m;j++){for(int k=i&(i-1);k;k=(k-1)&i)d[j][i]=min(d[j][i],d[j][i-k]+d[j][k]-weight[j]);if(d[j][i]<INF)Q.push(100000*j+i),in[j][i]=1;}while(!Q.empty()){int v=Q.front()/100000,t=Q.front()%100000;Q.pop();in[v][t]=0;for(int e=head[v];e!=-1;e=edge[e].next){int s=edge[e].to;if(d[s][t]>d[v][t]+edge[e].val+weight[s]-weight[v]){d[s][t]=d[v][t]+edge[e].val+weight[s]-weight[v];if(!in[s][t]){in[s][t]=1;Q.push(100000*s+t);}}}}}for(int i=1;i<(1<<n);i++)for(int j=0;j<n+m;j++)dp[i]=min(dp[i],d[j][i]);for(int i=1;i<(1<<n);i++){for(int j=i&(i-1);j;j=i&(j-1))dp[i]=min(dp[i],dp[j]+dp[i-j]);}cout<<dp[(1<<n)-1]<<endl;}return 0;
}