HDU 3311 Dig The Wells(spfa+ 압축 DP, 레벨 5)

H - Dig The Wells
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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Status
Appoint description:  System Crawler  (2013-05-30)
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
      
      
      
      

 
사고방식: 스탠나 나무, 먼저 전원 최단로를 만들고 DP[포함된 점집][뿌리 노드]를 사용한다.
            dp[mak][u]=MIN(dp[k][u]+dp[mask^k][u],dp[mask][j]+dis[j][u]);

///    
#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define ll(x) (1<<x)
using namespace std;
const int msize=1009;
const int oo=0x3f3f3f3f;
class Edge
{
  public:int v,w,next;
};
class Short_Path
{
public:
  int dis[msize][msize];
  bool vis[msize];
  int n,m,edge,head[msize];
  Edge e[msize*msize];
  void clear()
  {
    FOR(i,0,n+m)FOR(j,0,n+m)dis[i][j]=oo;
    edge=0;clr(head,-1);
  }
  void add(int u,int v,int w)
  {
    e[edge].v=v;e[edge].w=w;e[edge].next=head[u];head[u]=edge++;
  }
  void SPFA()
  {
    int kn=n+m,u,v;
    clr(vis,0);
    FOR(t,0,kn)
    {
      dis[t][t]=0;
      queue<int>Q;
      Q.push(t);
      while(!Q.empty())
      {
        u=Q.front();Q.pop();vis[u]=0;
        for(int i=head[u];~i;i=e[i].next)
        {
          v=e[i].v;
          if(dis[t][v]>dis[t][u]+e[i].w)
          {
            dis[t][v]=dis[t][u]+e[i].w;
            if(!vis[v])
            {
              Q.push(v);vis[v]=1;
            }
          }
        }
      }
    }
  }
  int dp[ll(7)][msize];///set contain , root
  void DP()
  { int kn=m+n;
    FOR(i,0,m+n)
    FOR(j,0,n)
    dp[ll(j)][i]=dis[i][j];///i->j
//    FOR(i,0,kn)FOR(j,0,kn)
//    printf("e=%d %d %d
",i,j,dis[i][j]);
    int mask=ll(n+1)-1;
    FOR(i,1,mask)
    if(i&(i-1))///   
    {
      FOR(j,0,kn)
      {
        dp[i][j]=oo;
        for(int k=i;k>0;k=(k-1)&i)
          dp[i][j]=min(dp[i][j],dp[k][j]+dp[i^k][j]);

      }
      FOR(j,0,kn)
      for(int k=0;k<=kn;++k)
        dp[i][j]=min(dp[i][j],dp[i][k]+dis[k][j]);
    }
    printf("%d
",dp[mask][0]);
  }
}sp;
int main()
{ int a,b,c,p;
  while(~scanf("%d%d%d",&sp.n,&sp.m,&p))
  {
    sp.clear();
    FOR(i,1,sp.n+sp.m)
    {
      scanf("%d",&a);
      sp.add(0,i,a);sp.add(i,0,a);
    }
    FOR(i,1,p)
    {
      scanf("%d%d%d",&a,&b,&c);
      sp.add(a,b,c);sp.add(b,a,c);
    }
    sp.SPFA();
    sp.DP();
  }
}