poj 3311(DP + 状态压缩)

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5205		Accepted: 2790

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
题目描述：
给定一系列点,求从0点出发，走完其它点，可以重复走，回到0点的最短路程.
假设n==4,那么点为0，1，2，3；
路径可能为0 1 2 3 0
        0 3 2 1 0
        0 3 1 2 0
     。。。。。。。 等等,
其实就是1到n-1的排列组合，每次有n-1中选择，需要算出每种状态的值，找到最小值，我们可以设到达j点，已经经过了s，s表示经过点的集合。
采用顺推的写法 dp[s | (1 <<k) ][k]=min(dp[s][j]+d[j][k],dp[s | (1<<k)][k]);
d数组为两点之间的最短路.

#include <stdio.h>
#include <cstring >
#include <cstdlib>
#include <iostream>
#define maxn 12
#define inf  0x3f3f3f3f
using namespace std;

int n;
int a[maxn][maxn];
int d[maxn][maxn];
int dp[<<maxn][maxn];
inline void  init()
{
   memset(d,inf,sizeof(d));
}
void floyed()
{
    for(int k=;k<n;k++)
     for(int i=;i<n;i++)
      for(int j=;j<n;j++)
    {
      d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
    }
}
void solve()
{
   /* for(int j=1;j<n;j++)
    {
       //dp[0][j]=d[0][j];
       dp[1][j]=d[0][j];

    }*/
    memset(dp,-,sizeof(dp));
    dp[][]=;
    for(int s=;s< ( << n);s++)
      for(int j=;j<n;j++)
      {
          if( dp[s][j]!=-)
          {
              for(int k=;k<n;k++)
              {
                if( (k!=j))
                {
                    if(dp[s | (<<k)][k]==- || (dp[s |(<<k)][k] > (dp[s][j]+d[j][k])))
                        dp[s | ( << k)][k]=dp[s][j]+d[j][k];
                   // cout<<dp[s | (1<<k )][k]<<endl;
                }

              }
          }
      }

    int ans=inf;
    int s=(<<n)-;
    /*for(int j=1;j<n;j++)
     {
         ans=min(ans,dp[s][j]+d[j][0]);
         printf("%d\n",dp[s][j]+d[j][0]);

     }*/
     printf("%d\n",dp[s][]);

}
int main()
{
    //cout<<(1<<2)<<endl;
  // freopen("test.txt","r",stdin);
   while(~scanf("%d",&n) && n!=)
   {
           init();
           n++;
       for(int i=;i<n;i++)
        for(int j=;j<n;j++)
      {
          scanf("%d",&a[i][j]);
          d[i][j]=a[i][j];
      }
      floyed();
      solve();
   }
    return ;
}