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; }
时间: 2024-08-24 03:06:08