POJ【3411】——Hie with the Pie

Time Limit: 2000MS	Memory Limit: 65536K

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

Source

East Central North America 2006

题意：给你n个地点和一个起点，要求每一个点都要至少走一次并且回到起点的距离最小。TSP问题我们可以预先处理出任意两点之间的距离，我们枚举中间的状态，那么初始的状态为dp[i][(1<< i )] = dis[0][i],然后枚举最后的点，进行转移。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

const int INF = 0x3f3f3f3f;

int mp[15][15];

int n;

int dp[15][500];

void GetFloyd()
{
    for(int i = 0; i<=n; i++)
    {
        for(int k = 0; k<=n; k++)
        {
            for(int s = 0; s<=n; s++)
            {
                mp[k][s] = min(mp[k][s],mp[k][i]+mp[i][s]);
            }
        }
    }
}

int main()
{
    while(~scanf("%d",&n) && n)
    {
        for(int i = 0; i<=n; i++)
        {
            for(int j = 0; j<=n; j++)
            {
                scanf("%d",&mp[i][j]);
            }
        }

        GetFloyd();

        for(int i =0; i<(1<<n); i++)//枚举状态
        {
            for(int j = 1; j <= n; j++)//枚举最后的一个点
            {
                if((1<<(j-1))&i)
                {
                    if((1<<(j-1)) == i)//初始化
                    {
                        dp[j][1<<(j-1)] = mp[0][j];
                    }
                    else
                    {
                        dp[j][i] = INF;

                        for(int k = 1; k<=n; k++)//枚举其他点，进行转移
                        {
                            if(((1<<(k-1))&i)&& k != j)
                            {
                                dp[j][i] = min(dp[j][i],dp[k][i^(1<<(j-1))]+mp[k][j]);
                            }
                        }
                    }
                }
            }
        }

        int ans = dp[1][(1<<n)-1]+mp[1][0];

        for(int i =2; i<=n; i++)
        {
            ans = min(ans,dp[i][(1<<n)-1]+mp[i][0]);
        }
        printf("%d\n",ans);
    }
    return 0;
}