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HDU 4770 Lights Against Dudely(暴力)
HDU 4770 Lights Against Dudely
题目链接
题意:给定灯,有一盏灯可以旋转,问最少几个灯可以照亮.的位置,并且不能照到#
思路:暴力求解,先枚举特殊的灯,再枚举正常的灯,要加剪枝,不然会TLE
代码:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; const int N = 205; const int d[4][3][2] = {{{0, 0}, {-1, 0}, {0, 1}}, {{0, 0}, {1, 0}, {0, 1}}, {{0, 0}, {1, 0}, {0, -1}}, {{0, 0}, {-1, 0}, {0, -1}}}; int n, m, g[N][N], x[N], y[N], pn, light[N][N], put[N][N], ans; char str[N]; bool judge(int x, int y, const int d[][2], int &ans) { if (put[x][y]) return false; ans = 0; for (int i = 0; i < 3; i++) { int xx = x + d[i][0]; int yy = y + d[i][1]; if (xx > n || xx < 1 || yy > m || yy < 1) continue; if (!g[xx][yy]) return false; } for (int i = 0; i < 3; i++) { int xx = x + d[i][0]; int yy = y + d[i][1]; if (xx > n || xx < 1 || yy > m || yy < 1) continue; if (light[xx][yy] == 0) ans++; light[xx][yy]++; } return true; } bool judge2(int now) { for (int i = 0; i < pn; i++) { if (x[i] <= now) break; if (!light[x[i]][y[i]]) return false; } return true; } int dfs(int now, int li, int num) { if (num > ans) return INF; if (!judge2(x[now])) return INF; if (li == 0) return num; if (now == pn) return INF; int tmp; int ans = INF; if (judge(x[now], y[now], d[0], tmp)) { ans = min(ans, dfs(now + 1, li - tmp, num + 1)); for (int i = 0; i < 3; i++) { int xx = x[now] + d[0][i][0]; int yy = y[now] + d[0][i][1]; if (xx > n || xx < 1 || yy > m || yy < 1) continue; light[xx][yy]--; } } ans = min(ans, dfs(now + 1, li, num)); return ans; } int main() { while (~scanf("%d%d", &n, &m) && n || m) { for (int i = 1; i <= n; i++) { scanf("%s", str + 1); for (int j = 1; j <= m; j++) g[i][j] = (str[j] == '.'?1:0); } pn = 0; for (int i = n; i >= 1; i--) { for (int j = 1; j <= m; j++) { if (g[i][j]) { x[pn] = i; y[pn] = j; pn++; } } } memset(light, 0, sizeof(light)); ans = INF; ans = dfs(0, pn, 0); for (int i = 0; i < pn; i++) { for (int j = 0; j < 4; j++) { int tmp = 0; memset(light, 0, sizeof(light)); if (judge(x[i], y[i], d[j], tmp)) { put[x[i]][y[i]] = 1; ans = min(ans, dfs(0, pn - tmp, 1)); put[x[i]][y[i]] = 0; } } } if (ans == INF) printf("-1\n"); else printf("%d\n", ans); } return 0; }
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