【状压DP】Hie with the Pie

poj3311

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.
给你一个有n+1(1<=n<=10)个点的有向完全图，用矩阵的形式给出任意两个不同点之间的距离。（其中从i到j的距离不一定等于从j到i的距离）现在要你求出从0号点出发，走过1到n号点至少一次，然后再回到0号点所花的最小时间

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

Solution

这道题和另外一道 售货员的难题 挺像的，只不过相对于那道题，此题需要进行一些转换

可以发现题目给出的距离是任意两点间的距离，但这个距离并不是任意两点间的最短距离，所以我们先跑一个最短路处理出任意两点间的最短距离

直接暴搜状态数巨大，考虑状压DP

状态设计成二维，dp[ i ][ j ]表示到达 i 号节点时，节点访问情况为 j 的最小时间（j 为二进制状压）

状态转移方程

dp[i][m | (1 << i)] = min(dp[i][m | (1 << i)], dp[j][m] + dis[j][i]);

关于状压DP可参考本人博客 链接

Code

#include<bits/stdc++.h>
using namespace std;

int n;
int dis[20][20];
int dp[20][(1 << 11) + 7];
int ans;

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
    return x * f;
}


int main() {
    while (true) {
	memset(dis, 0x3f3f3f3f, sizeof dis);
	memset(dp, 0x3f3f3f3f, sizeof dp);
	n = read() + 1;
	if (n == 1) break;
	for (int i = 0; i < n; i++) {
	    for (int j = 0; j < n; j++) {
		dis[i][j] = read();
	    }
	}
	//floyed
	for (int k = 0; k < n; k++) {
	    for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
		    dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
//		    cout << dis[i][j] << endl;
		}
	    }
	}
	//dp
	for (int m = 0; m < (1 << n); m++) {
	    for (int i = 0; i < n; i++) {//next
		if (m & (1 << i)) continue;
		dp[i][(1 << i)] = dis[0][i];
                for (int j = 0; j < n; j++) {//now
		    if (m & (1 << j)) {
			dp[i][m | (1 << i)] = min(dp[i][m | (1 << i)], dp[j][m] + dis[j][i]);
		    }
		}
	    }
	}
	ans = 0x3f3f3f3f;
	for (int i = 1; i < n; i++) {
	    ans = min(ans, dp[i][(1 << n) - 2] + dis[i][0]);
	}
	printf("%d\n", ans);
    }
}