Sha 10798-Be wary of Roses (memory-based BFS)

Sha 10798-Be wary of Roses (memory-based BFS)
Ultraviolet A 10798-Be wary of Roses

Question Link

Given a map, people start at the center and choose a method to go out to minimize the maximum value of the flowers they step in any direction.

Train of Thought: Use the priority queue for BFS. When the minimum step is taken out each time, the status of the extensive search record is the current position, and the number of flowers in four directions.

Code:

#include 
 
  #include 
  
   #include #include 
   
    using namespace std;const int N = 21;const int d[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};int n, vis[N][N][11][11][11][11];char g[N][N];struct State {int x, y, val;int up, left, down, right;State() {x = y = up = left = down = right = 0;}State(int x, int y, int up, int left, int down, int right) {this->x = x;this->y = y;this->up = up;this->left = left;this->down = down;this->right = right;val = max(max(max(up,left), down), right);}bool operator < (const State& c) const {return val > c.val;}} s;void init() {for (int i = 0; i < n; i++) {scanf("%s", g[i]);for (int j = 0; j < n; j++)if (g[i][j] == 'P')s.x = i, s.y = j;}}int bfs() {memset(vis, 0, sizeof(vis));priority_queue
    
      Q;Q.push(s);vis[s.x][s.y][0][0][0][0] = 1;while (!Q.empty()) {State u = Q.top();Q.pop();if (u.x == 0 || u.x == n - 1 || u.y == 0 || u.y == n - 1) return u.val;for (int i = 0; i < 4; i++) {int xx = u.x + d[i][0];int yy = u.y + d[i][1];int up = u.up;int left = u.left;int down = u.down;int right = u.right;if (g[xx][yy] == 'R') up++;if (g[n - 1 - yy][xx] == 'R') left++;if (g[n - 1 - xx][n - 1 - yy] == 'R') down++;if (g[yy][n - 1 - xx] == 'R') right++;if (!vis[xx][yy][up][left][down][right]) {vis[xx][yy][up][left][down][right] = 1;Q.push(State(xx, yy, up, left, down, right));}}}}int main() {while (~scanf("%d", &n) && n) {init();printf("At most %d rose(s) trampled.\n", bfs());}return 0;}