POJ 2044 Weather Forecast

来源：互联网发布：淘宝卡通行走人偶编辑：程序博客网时间：2024/05/16 07:56

题意：有一朵2*2的云朵，和一个4*4的地区。被云层覆盖的区域在当天一定有雨下，云层有4种移动方式。但是规定在城市或者节日期间希望不要下雨，而且一个地方不能有连续7天没下雨。

思路：首先解决一个地方不能有连续7天没下雨的情况,要让地图上的所有地方都覆盖到的话，只要4个角都覆盖到的话，就行了，因为要到那里就要经过所有的了，然后就是状态的记录了，vis[k][sign][a][b][c][d]，表示第k天点是sign的四个格子的下雨情况

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node {int a,b,c,d;node():a(0), b(0), c(0), d(0) {}};struct point {int x,y;point(int _x, int _y):x(_x), y(_y) {}};const int dir[9][2] = {{0, 0}, {-1, 0}, {-2, 0}, {0, -1},{0, -2}, {1, 0}, {2, 0}, {0, 1}, {0, 2}};int vis[370][9][7][7][7][7];int day[370], n;int getFlag(int i) {return 1<<i;}int check(int k, const point &u, const node &s) {if (s.a == 7 || s.b == 7 || s.c == 7 || s.d == 7)return 0;int flag = 0;int x = u.x, y = u.y;flag |= getFlag(4*x+y) | getFlag(4*x+y+1);flag |= getFlag(4*(x+1)+y) | getFlag(4*(x+1)+y+1);if (flag & day[k])return 0;if (vis[k][4*x+y][s.a][s.b][s.c][s.d])return 0;vis[k][4*x+y][s.a][s.b][s.c][s.d] = 1;return 1;}int dfs(int k, const point &u, const node &state) {if (k == n)return 1;node s = state;s.a += 1, s.b += 1;s.c += 1, s.d += 1;if (u.x == 0 && u.y == 0)s.a = 0;else if (u.x == 0 && u.y == 2)s.b = 0;else if (u.x == 2 && u.y == 0)s.c = 0;else if (u.x == 2 && u.y == 2)s.d = 0;if (!check(k, u, s)) return 0;for (int i = 0; i < 9; i++) {int nx = u.x + dir[i][0];int ny = u.y + dir[i][1];if (nx >= 0 && nx < 3 && ny >= 0 && ny < 3)if (dfs(k+1, point(nx, ny), s))return 1;}return 0;}int main() {while (scanf("%d", &n) != EOF && n) {for (int i = 0; i < n; i++) {day[i] = 0;for (int j = 0; j < 16; j++) {int x;scanf("%d", &x);day[i] <<= 1;day[i] |= x;}memset(vis[i], 0, sizeof(vis[i]));}node s;if (dfs(0, point(1, 1), s))printf("1\n");else printf("0\n");}return 0;}

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