POJ 3311 Hie with the Pie（状态压缩DP）

Hie with the Pie

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6776 Accepted: 3652

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). Thejth 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

30 1 10 101 0 1 210 1 0 1010 2 10 00

Sample Output

8

Source

East Central North America 2006

大体题意：外卖员要去送比萨，给出外卖店与所有顾客两两之间的距离，求将所有比萨送出并返回比萨店的最短时间。

大体思路：首先用Floyd求一遍最短距离。因为顾客数很少，可以用一个二进制串来表示是否给第i名顾客送过（0表示没有，1表示有）。我们用DP（S，i）表示“在给第j位顾客送过比萨后状态为S”，那么DP的边界就为DP（1 << i,  i）= dis[0][i]，即只给第i位顾客送去比萨的最短距离为从比萨店到该顾客的距离。状态转移如下：如果状态S中有第i位顾客，那么给第i位顾客送餐的最短距离为（先给第j位顾客送（这个状态要有j不能有i），再从j到i的最短距离，相当于一个中转）。最后的答案就是MIN{ DP( (1 << n) - 1, i) + dis[i][0]}。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int INF = 1e9 + 10;const int maxn = 15;int dp[1<<11][maxn], dis[maxn][maxn];int main(){    int n;    while(scanf("%d",&n) == 1 && n) {        for (int i = 0; i <= n; i++)          for (int j = 0; j <= n; j++)             scanf("%d",&dis[i][j]);                     //Floyd        for (int k = 0; k <= n; k++)          for (int i = 0; i <= n; i++)            for (int j = 0; j <= n; j++)               dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);                       //DP求解        for (int S = 1; S < (1<<n); S++) {            for (int i = 1; i <= n; i++) if (S & (1<<(i-1))) { //如果状态中有第i名顾客                if (S == (1<<(i-1))) dp[S][i] = dis[0][i]; //DP边界，如果只给第i名顾客送，那么距离就是起点到这位顾客的距离。                else {                    dp[S][i] = INF;                    for (int j = 1; j <= n; j++) if (S & (1<<(j-1))) { //先给j送，再从j到i                        if (j == i) continue;                        dp[S][i] = min(dp[S][i], dp[S^(1<<(i-1))][j] + dis[j][i]);                    }                }            }        }                int mi = INF;        for (int i = 1; i <= n; i++) {            mi = min(mi, dp[(1<<n)-1][i] + dis[i][0]);        }        printf("%d\n",mi);    }}