POJ3311 Hie with the Pie 【状压dp/TSP问题】

题目链接：http://poj.org/problem?id=3311

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:12225		Accepted: 6441

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.

题目大意：从 0 点出发，经过其他 n 点（允许重复经过），最后回到 0 点，要求输出最短距离。

思路：

1.属于经典的旅行商问题（TSP问题）：从一个点出发可重复的经过其他至少一次，最后回到该点，问如何选择路线使得路径最短。TSP问题需要用到状压dp来记录状态，这里有一篇博客：TSP问题总结归纳以及例题

2.还需要先跑一遍floyd得到任意两点之间的最短距离。因为图中两点之间并不一定是最短的路径。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define mem(a, b) memset(a, b, sizeof(a))
#define LL long long
using namespace std;
const int MAXN = 15;
const int inf = 0x3f3f3f3f;

int n;
LL map[MAXN][MAXN], dis[MAXN][MAXN];
LL dp[1 << 11][MAXN]; //表示达到 i 状态时，最后访问的位置是 j  

void floyd()
{
    for(int i = 0; i <= n; i ++)
        for(int j = 0; j <= n; j ++)
            for(int k = 0; k <= n; k ++)
                dis[i][j] = min(dis[i][j], map[i][k] + map[k][j]);
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0)
            break;
        for(int i = 0; i <= n; i ++)
        {
            for(int j = 0; j <= n; j ++)
            {
                scanf("%lld", &map[i][j]);
            }
        }
        mem(dis, inf);
        floyd(); //处理出任意两点之间的最短距离
        for(int i = 0; i <= ((1 << n) - 1); i ++)//枚举状态 
        {
            int S = i; //state
            for(int j = 1; j <= n; j ++)
            {
                if(S == (1 << (j - 1))) //只经过 j 这个点 
                    dp[S][j] = dis[0][j];
                else
                {
                    dp[S][j] = inf;
                    for(int k = 1; k <= n; k ++) //与 j 不同的点
                    {
                        if(k != j && (S & (1 << (k - 1))))
                            dp[S][j] = min(dp[S][j], dp[S ^ (1 << (j - 1))][k] + dis[k][j]);
                    }
                }
            }
        }
        LL ans = dp[(1 << n) - 1][1] + dis[1][0];
        for(int i = 2; i <= n; i ++)
            ans = min(ans, dp[(1 << n) - 1][i] + dis[i][0]);
        printf("%lld
", ans);
    }
    return 0;
}