POJ3311 ---Hie with the Pie（状压dp）

题目

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

大意

给一个图，有n+1个点，每两个点之间都有一个距离，求从0号节点遍历每一个节点最后回到0号节点的最短距离（每个点可以经过多次）

一道标准的状压dp了吧，$dp[state][i]$表示当前状态为state，正处在点i时的最短路径；只需要枚举state中已经经过的中间点，取最小的就可以了
任意两点之间的距离直接用floyd就可以了

#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<string.h>
using namespace std;
int dis[12][12],dp[1<<12][12],n;
inline int read(){
    char ch=getchar();
    int res=0;
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
    return res;
}
inline void floyd()//floyd 求最短路 
{
    for(int k=0;k<=n;k++)
    {
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                if(dis[i][j]>dis[i][k]+dis[k][j])
                dis[i][j]=dis[i][k]+dis[k][j];
            }
        }
    }
}
int main(){
    n=read();
    while(n)
    {
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                dis[i][j]=read();
            }
        }
        floyd();
        int state=(1<<n)-1;//所有状态 
        for(int s=0;s<=state;s++)
        {
            for(int i=1;i<=n;i++)
            {
                if(s&(1<<(i-1)))//如果已经经过这个点 
                {
                    if(s==(1<<(i-1)))
                    {
                        dp[s][i]=dis[0][i];//如果目前只经过这个点，那dp的值就是这个点到原点的距离 
                    }
                    else 
                    {
                        dp[s][i]=1<<30;
                        for(int j=1;j<=n;j++)//枚举中间点求最小值 
                        {
                            if(s&(1<<(j-1))&&j!=i)
                            {
                                dp[s][i]=min(dp[s][i],dp[s^(1<<(i-1))][j]+dis[j][i]);
                            }
                        }
                    }
                }
            }
        }
        int mina=dp[(1<<n)-1][1]+dis[1][0];//最后加上从目前该点回到0点的时间 
        for(int i=2;i<=n;i++)
        {
            mina=min(dp[state][i]+dis[i][0],mina);
        }
        cout<<mina<<'\n';
        n=read();//有多组数据 
    }
}