Hdu5094Maze (BFS)

Link to the question: hdu5094Maze: Given a graph, you need to move from () to (n, m), and some of them have doors or walls in the middle. The corresponding door must have the corresponding key to pass through. And specify the location of the key. Solution: it is a common bfs. the ownership of the key can be represented by a binary number. The only trouble is that the door and wall are on the edge, so I am working on the edge directly during preprocessing. #

Link: hdu 5094 Maze

For a given graph, we need to move from () to (n, m). some of them have doors or walls in the middle. The corresponding door needs to be matched

. And specify the location of the key.

Solution: it is a common bfs. the ownership of the key can be represented by a binary number. The only trouble is that the door and wall are on the edge.

It is processed directly on the edge.

#include 
  
   #include 
   
    #include 
    
     #include 
     
      #include using namespace std;typedef pair
      
        pii;const int maxn = 55;const int maxs = 3000;const int inf = 0x3f3f3f3f;const int dir[4][2] = {{-1, 0}, {0, 1}, {0, -1}, {1, 0}};int N, M, P, v[maxn][maxn];int g[maxn * maxn][4], dp[maxn][maxn][maxs];inline int idx(int x, int y) { return (x-1) * M + y - 1;}void init () { memset(v, 0, sizeof(v)); memset(g, -1, sizeof(g)); memset(dp, inf, sizeof(dp)); int n, x1, y1, x2, y2, k, d; scanf("%d", &n); while (n--) { scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &k); if (x1 == x2) d = (y1 > y2 ? 2 : 1); else d = (x1 > x2 ? 0 : 3); g[idx(x1, y1)][d] = g[idx(x2, y2)][3-d] = k; } scanf("%d", &n); while (n--) { scanf("%d%d%d", &x1, &y1, &k); v[x1][y1] |= (1<<(k-1)); }}int bfs () { queue
       
         Q; Q.push(make_pair(idx(1, 1), 0)); dp[1][1][0] = 0; while (!Q.empty()) { pii u = Q.front(); Q.pop(); int x = u.first / M + 1; int y = u.first % M + 1; if (x == N && y == M) return dp[x][y][u.second]; for (int i = 0; i < 4; i++) { if (g[u.first][i] == 0) continue; int xx = x + dir[i][0]; int yy = y + dir[i][1]; if (xx <= 0 || xx > N || yy <= 0 || y > M) continue; int s = u.second | v[xx][yy]; int tmp = g[u.first][i]; if (tmp == -1 || (u.second & (1<<(tmp-1)))) { if (dp[xx][yy][s] > dp[x][y][u.second] + 1) { dp[xx][yy][s] = dp[x][y][u.second] + 1; Q.push(make_pair(idx(xx, yy), s)); } } } } return -1;}int main () { while (scanf("%d%d%d", &N, &M, &P) == 3) { init(); printf("%d\n", bfs()); } return 0;}