poj 3311 Hie with the Pie dp+状压

Hie with the Pie

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4671		Accepted: 2471

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

题意

从0出发送汉堡，送完每一个点后回到0点

矩阵 表示各个点间距离。

思路

由于能够各个点多次经过。所以先求下floyd 来更新点点之间最短距离。

把每一步。到各个点的状态和最后一步所在的位置还有所花的距离保存下来。

把最后到的各个地方，再加个回零点的距离。求个最小值。

dp[15][15]   一维表示已经走的步数，二维表示当前最后一步到的点。  map的x表示状态，y表示已经花费的时间。

#include<stdio.h>
#include<map>
#include<algorithm>
using namespace std;
map<int,int>::iterator it;
map<int,int> dp[15][15];//ceng  结尾
int mp[15][15];
void floyd(int n)
{
	int i,j,k;
	for(k=0;k<=n;k++)
	{
		for(i=0;i<=n;i++)
		{
			for(j=0;j<=n;j++)
			{
				mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
			}
		}
	}
}
int main()
{
	int n,i,j,k,mn,x,y;
	while(scanf("%d",&n),n)
	{
		for(i=0;i<=n;i++)
		{
			for(j=0;j<=n;j++)
			{
				scanf("%d",&mp[i][j]);
			}
		}
		floyd(n);
		/*for(i=0;i<=n;i++)
		{
			for(j=0;j<=n;j++)
			{
				printf("%d ",mp[i][j]);
			}
			puts("");
		}*/
 
		//clear
		dp[0][0][1]=0;
		for(i=1;i<=n;i++)//第几处了
		{
			for(k=0;k<=n;k++)//终点
			{
				dp[i][k].clear();
				for(j=0;j<=n;j++)//起点
				{
					for(it=dp[i-1][j].begin();it!=dp[i-1][j].end();it++)//上一层的各个状态
					{
						x=it->first;
						y=it->second;
						if((x&(1<<k))==0)//还没走过
						{
							if(dp[i][k].count(x|(1<<k))==0)
								dp[i][k][x|(1<<k)]=y+mp[j][k];
							else
							dp[i][k][x|(1<<k)]=min(dp[i][k][x|(1<<k)],y+mp[j][k]);
						}
					}
				}
			}
		}
 
		i=n;//最后一层
		mn=999999999;
		for(j=1;j<=n;j++)
		{
			for(it=dp[i][j].begin();it!=dp[i][j].end();it++)//上一层的各个状态
			{
				mn=min(mn,it->second+mp[j][0]);
			}
		}
		printf("%d\n",mn);
 
	}
	return 0;
}
/*
3
0 1 10 10
1 0 1 2
10 1 0 10
10 2 10 0
 
3
0 5 6 7
5 0 8 9
6 8 0 10
7 9 10 0
*/