<状压DP>solution-POJ3311_Hie with the Pie

Hie with the Pie

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

人话：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

首先，先把每两个点之间的最短路求出来，使用floyd搞定

然后开始状压dp

我们把当前去过哪些点进行状压，i二进制表示从左到右第k位表示第k个点是否访问过

那么我们就可以把dp数组搞出来了，\(f[i][j]\)表示在已访问i状态这么多点的情况下，重点是j的最短路

状态转移方程就是：

\(f[i|(1<<k)][k] = f[i][j]+dis[j][k]\)

还有，起始点状态记得初始化

Code:

#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#include <iostream>
#define reg register
using namespace std;
const int MaxN=11;
const int inf=0x3f3f3f3f;
template <class t> inline void rd(t &s)
{
	s=0;
	reg char c=getchar();
	while(!isdigit(c))
		c=getchar();
	while(isdigit(c))
		s=(s<<3)+(s<<1)+(c^48),c=getchar();
	return;
}
int n;
int f[(1<<MaxN)+1][MaxN];
int G[MaxN][MaxN],dis[MaxN][MaxN];
inline void work()
{
	memset(f,0x3f,sizeof f);f[1][0]=0;
	for(int i=0;i<=n;++i)
		for(int j=0;j<=n;++j)
			rd(G[i][j]),dis[i][j]=G[i][j];
	for(int k=0;k<=n;++k)
		for(int i=0;i<=n;++i) if(k!=i)
			for(int j=0;j<=n;++j) if(i!=j)
				dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
	for(int i=1;i<(1<<(n+1));++i)
	{
		for(int j=0;j<=n;++j) if(f[i][j]!=inf)
			for(int k=0;k<=n;++k)
				if(j!=k)
					f[i|(1<<k)][k]=min(f[i|(1<<k)][k],f[i][j]+dis[j][k]);
	}
	reg int u=(1<<(n+1))-1;
	printf("%d\n",f[u][0]);
	return;
}
signed main(void)
{
	while(cin>>n&&n)
		work();
	return 0;
}