Hie with the Pie - POJ 3311 状压dp

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3819		Accepted: 2009

Description

The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.

Input

Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting any other locations along the way. Note that there may be quicker ways to go from i to j via other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.

Output

For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.

Sample Input

3
0 1 10 10
1 0 1 2
10 1 0 10
10 2 10 0
0

Sample Output

8

题意：一个人去送披萨，给你每两个点之间的距离，并且最后回到披萨店，问你最短路程是多少。注意两个点之间来回的距离可以不一样。

思路：首先用Floyd算出没两个点之间的最短距离，然后用状压dp，dp[S][i]表示他经过某些点后并且在i点的最短距离。

AC代码如下;

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dis[20][20],dp[2000][20];
int main()
{ int n,i,j,k,S,ans;
  while(~scanf("%d",&n) && n>0)
  { memset(dp,0,sizeof(dp));
    for(i=0;i<=n;i++)
     for(j=0;j<=n;j++)
      scanf("%d",&dis[i][j]);
    for(k=0;k<=n;k++)
     for(i=0;i<=n;i++)
      for(j=0;j<=n;j++)
       dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
    for(S=0;S<(1<<n);S++)
     for(i=1;i<=n;i++)
      if(S&(1<<(i-1)))
       if(S == (1<<(i-1)))
        dp[S][i]=dis[0][i];
       else
       { dp[S][i]=1000000000;
         for(j=1;j<=n;j++)
          if((S&(1<<(j-1))) && i!=j)
           dp[S][i]=min(dp[S][i],dp[S^(1<<(i-1))][j]+dis[j][i]);
       }
    S=(1<<n)-1;
    ans=dp[S][1]+dis[1][0];
    for(i=2;i<=n;i++)
     ans=min(ans,dp[S][i]+dis[i][0]);
    printf("%d\n",ans);
  }
}