POJ3311——Hie with the Pie（状态压缩DP）

Hie with the Pie

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 2917Accepted: 1492

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

解析：

          dp[i][s]表示走到第i个点，状态为s，其中，经过的点用二进制1表示，没经过用二进制0表示。。。

          先用弗洛伊德预处理出任意点间的最短路径。。。

         看了大神的代码—>http://blog.csdn.net/accry/article/details/6607703

代码：

#include<cstdio>#include<algorithm>using namespace std;#define inf 100000000int n,dis[11][11],dp[11][1<<11];void read(){    freopen("poj3311.in","r",stdin);    freopen("poj3311.out","w",stdout);}void work(){    while(scanf("%d",&n)&&n)    {        for(int i=0;i<=n;i++)            for(int j=0;j<=n;j++)            scanf("%d",&dis[i][j]);        for(int k=0;k<=n;k++)//floyd预处理任意两点之间的最短路            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];       for(int s=0;s<(1<<n);s++)           for(int i=1;i<=n;i++)           {                if(s&(1<<(i-1)))//状态s已经过城市i                {                    if(s==(1<<(i-1)))dp[i][s]=dis[0][i];//只经过i城市，最短距离就是dis[0][i]                    else                    {                        dp[i][s]=inf;                        for(int k=1;k<=n;k++)                            if(s&(1<<(k-1)) && k!=i)//再枚举一个k来更新dis值                                dp[i][s]=min(dp[k][s^(1<<(i-1))]+dis[k][i],dp[i][s]);                    }                }            }        int ans=inf;        for(int i=1;i<=n;i++)        if(dp[i][(1<<n)-1]+dis[i][0]<ans)        ans=dp[i][(1<<n)-1]+dis[i][0];        printf("%d\n",ans);    }}int main(){    read();    work();    return 0;}