POJ3311 Hie with the Pie

http://poj.org/problem?id=3311

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.

题意：编号为0~n的点，要求从0出发，走完所有1~n点，最后回到0的最小总路程。可以多次经过某些点。

思路：先floyd预处理任意两点最短路长度，然后就是裸的tsp问题的。设f(i,s):当前在i点，当前对0~n个点的是否访问的压位集合，对应的最短路程。

Q：为什么只求两点最短路长度，不用存储这个最短路经过了哪些点？

A：因为这是dp，把所有的路径组合都遍历过了。如果A->B->C是A->C的最优，有A->C的决策，也有先A->B，再B->C的决策，不可能会漏解。

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

int n,d[15][15],cover[15][15],f[15][1<<15];

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

void dp()
{
	memset(f,0x3f,sizeof(f));
	f[0][(1<<(n+1))-1]=0;
	for(int s=(1<<(n+1))-2;s>=0;s--)
	{
		for(int u=0;u<=n;u++)
		{
			if((s&(1<<u) || !s&&!u))
			for(int v=0;v<=n;v++)if(!(s&(1<<v)))
			{
				f[u][s]=min(f[u][s],f[v][s|(1<<v)]+d[u][v]);
			}
		}
	}
}

int main()
{
	//freopen("input.in","r",stdin);
	while(cin>>n&&n)
	{
		for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)cin>>d[i][j];
		floyd();
		dp();
		cout<<f[0][0]<<endl;
	}
	return 0;
}