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HDU 5067 Harry And Dig Machine

思路：由于点才10个，在加上一个起点，处理出每个点之间的曼哈顿距离，然后用状压dp搞，状态表示为：
dp[i][s]，表示在i位置，走过的点集合为s的最小代价

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int N = 15;
int n, m;

struct Point {
	int x, y;
	Point() {}
	Point(int x, int y) {
		this->x = x;
		this->y = y;
	}
} p[N];

int pn;

int g[N][N];
int dp[N][(1<<13) + 5];

int dis(Point a, Point b) {
	return abs(a.x - b.x) + abs(a.y - b.y);
}

const int INF = 0x3f3f3f3f;
int main() {
	while (~scanf("%d%d", &n, &m)) {
		pn = 0;
		int a;
		int flag = 0;
		int zero;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				scanf("%d", &a);
				if (a) {
					p[pn++] = Point(i, j);
					if (i == 0 && j == 0) {
						flag = 1;
						zero = pn - 1;
					}
				}
			}
		}
		if (!flag) {
			zero = pn;
			p[pn++] = Point(0, 0);
		}
		for (int i = 0; i < pn; i++) {
			for (int j = i; j < pn; j++) {
				g[i][j] = g[j][i] = dis(p[i], p[j]);
			}
		}
		for (int i = 0; i < pn; i++)
			for (int j = 0; j < (1<<pn); j++)
				dp[i][j] = INF;

		dp[zero][0] = 0;
		dp[zero][(1<<zero)] = 0;
		int ss = (1<<pn);
		for (int i = 0; i < ss; i++) {
			for (int j = 0; j < pn; j++) {
				if (i&(1<<j)) {
					for (int k = 0; k < pn; k++) {
						if (i&(1<<k)) {
							dp[j][i] = min(dp[j][i], dp[k][i^(1<<j)] + g[j][k]);
						}
					}
				}
			}
		}
		int ans = INF;
		for (int i = 0; i < pn; i++)
			ans = min(ans, dp[i][(1<<pn) - 1] + g[zero][i]);
		printf("%d\n", ans);
	}
	return 0;
}

HDU 5067 Harry And Dig Machine（状压dp）

标签：style   blog   color   io   os   ar   for   sp   2014   

原文地址：http://blog.csdn.net/accelerator_/article/details/40221345