poj3311（状态压缩dp）

Hie with the Pie

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5896 Accepted: 3180

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

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

Sample Output

8

题意：输入一个数n，现在有n个地方（标号1到n）要从标号为0的地方出去，经过所有的地方之后回来，求最短的时间，输入(n+1)*(n+1)的矩阵表示每两点之间到达所需要的时间。

思路：用dp[i][j]表示状态i下以j结尾的最短时间。由于不知道每两个点之间到底怎么走时间最短，用floyd求一下最短路程，然后进行dp。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;typedef long long ll;int dp[1<<11][11];///dp[i][j]表示i状态下以j结尾的最短时间int tu[15][15];int n;const int INF=99999999;void floyd()///floyd求两点间的最短路{    for(int k=0; k<n; k++)        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                tu[i][j]=min(tu[i][j],tu[i][k]+tu[k][j]);}int main(){    while(~scanf("%d",&n)&&n++)    {        for(int i=0; i<(1<<n); i++)            for(int j=0; j<n; j++)                dp[i][j]=INF;        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                scanf("%d",&tu[i][j]);        floyd();        for(int i=1; i<n; i++)            dp[1<<i][i]=tu[0][i];        for(int j=0; j<1<<n; j++)            for(int i=0; i<n; i++)                if(dp[j][i]!=INF)                    for(int k=1; k<n; k++)                        if(!(j&(1<<k)))                            dp[j|(1<<k)][k]=min(dp[j][i]+tu[i][k],dp[j|(1<<k)][k]);        int ans=INF;        for(int i=1; i<n; i++)            ans=min(dp[(1<<n)-2][i]+tu[i][0],ans);        cout<<ans<<endl;    }    return 0;}