Hie with the Pie 状压+最短路

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

思路：最短路求出每两点之间的最短距离，然后将每个的状态压缩，1表示这个点已经去过，0表示这个点没有去过

dp[i][j] = min(dp[i-(1<<j)][k] + dis[k][j] ,dp[i][j]) dp[i][j] 表示 i 状态 到 j 的最短距离

i 表示 状态，j,k是这个状态里已经走过的点，这样就可以用当前值和 这个状态还没有走j的前一个状态加上从k到j的距离进行比较。

最后 dp[(1<<n)-1][0] 所有点都走过的这个状态最后返回起点的最短距离

AC代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<string>
#include<map>
#include<cmath>
#define INF 0x3f3f3f3f
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fori(x) for(int i=0;i<x;i++)
#define forj(x) for(int j=0;j<x;j++)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define sca(x) scanf("%d", &x)
#define scas(x) scanf("%s",x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pris(x) printf("%s\n",x)
#define prl(x) printf("%lld\n",x)
//#include <bits/stdc++.h>

typedef long long ll;
const int maxn=1e6+7;
const int mod=1e9+7;
const double eps=1e-8;
// const double pi = acos(-1);

using namespace std;

int dis[100][100];
int cost[100][100];
int dp[1<<11][11];
int main()
{
    int n;
    while(sca(n) && n)
    {
      n++;
      rep(i,0,n)
      rep(j,0,n)
      {
        sca(cost[i][j]);
        dis[i][j] = cost[i][j];
      }
      rep(i,0,n)
      {
        rep(j,0,n)
        {
          rep(k,0,n)
          {
            if(dis[i][j] > dis[i][k] + dis[k][j])
            {
              dis[i][j] = dis[i][k] + dis[k][j];
            }
          }
        }
      }
      memset(dp,INF);
      dp[1][0] = 0;
      rep(i,1,(1<<n))
      {
        rep(j,0,n)
        {
          if(i&(1<<j))
          {
            if(i - (1<<j) == 0)
            {
              dp[i][j] = dis[0][j];
              continue;
            }
            int temp = i - (1<<j);
            rep(k,0,n)
            {
              if((1<<k)&temp)
              {
                if(dp[i][j] > dp[temp][k] + dis[k][j])
                {
                  dp[i][j] = dp[temp][k] + dis[k][j];
                }
              }
            }
          }
        }
      }
      printf("%d\n",dp[(1<<n)-1][0]);    
    }
    return 0;
}