hdu 1429 bfs+狀態壓縮

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對於這類需要鑰匙才能通過的迷宮來說，一般visit數組會開成三維的：bool visit[N][N][1 << 10];最後一維表示通過該位置時的擁有的鑰匙的情況。

  1 #include <iostream>  2 #include <cstring>  3 #include <cstdio>  4 #include <cctype>  5 #include <queue>  6 using namespace std;  7   8 const int N = 21;  9 char maze[N][N]; 10 bool visit[N][N][1 << 10]; 11 int dx[] = { 0, 0, 1, -1 }; 12 int dy[] = { 1, -1, 0, 0 }; 13 int sx, sy, ex, ey; 14 int n, m, t; 15  16 bool ok( int x, int y ) 17 { 18     return x >= 0 && x < n && y >= 0 && y < m; 19 } 20  21 struct Node  22 { 23     int x, y, step, state; 24     Node(){} 25     Node( int _x, int _y, int _step, int _state ) 26     { 27         x = _x, y = _y, step = _step, state = _state; 28     } 29 }; 30  31 queue<Node> q; 32  33 int bfs() 34 { 35     while ( !q.empty() ) q.pop(); 36     memset( visit, false, sizeof(visit) ); 37     q.push( Node( sx, sy, 0, 0 ) ); 38     visit[sx][sy][0] = true; 39     while ( !q.empty() ) 40     { 41         Node cur = q.front(); 42         q.pop(); 43         if ( cur.step == t ) return -1; 44         if ( cur.x == ex && cur.y == ey ) return cur.step; 45         for ( int i = 0; i < 4; i++ ) 46         { 47             int xx = cur.x + dx[i]; 48             int yy = cur.y + dy[i]; 49             if ( ok( xx, yy ) ) 50             { 51                 if ( maze[xx][yy] == ‘*‘ ) continue; 52                 if ( isupper( maze[xx][yy] ) ) 53                 { 54                     int tmp = ( 1 << ( maze[xx][yy] - ‘A‘ ) ); 55                     if ( ( cur.state & tmp ) && !visit[xx][yy][cur.state] ) 56                     { 57                         q.push( Node( xx, yy, cur.step + 1, cur.state) ); 58                         visit[xx][yy][cur.state] = true; 59                     } 60                 } 61                 else if ( islower( maze[xx][yy] ) ) 62                 { 63                     int tmp = ( 1 << ( maze[xx][yy] - ‘a‘ ) ); 64                     if ( !visit[xx][yy][cur.state | tmp] ) 65                     { 66                         q.push( Node( xx, yy, cur.step + 1, cur.state | tmp ) ); 67                         visit[xx][yy][cur.state | tmp] = true; 68                     } 69                 } 70                 else 71                 { 72                     if ( !visit[xx][yy][cur.state] ) 73                     { 74                         q.push( Node( xx, yy, cur.step + 1, cur.state ) ); 75                         visit[xx][yy][cur.state] = true; 76                     } 77                 } 78             } 79         } 80     } 81     return -1; 82 } 83  84 int main () 85 { 86     while ( scanf("%d%d%d", &n, &m, &t) != EOF ) 87     { 88         for ( int i = 0; i < n; i++ ) 89         { 90             scanf("%s", maze[i]); 91             for ( int j = 0; j < m; j++ ) 92             { 93                 if ( maze[i][j] == ‘@‘ ) 94                 { 95                     sx = i; 96                     sy = j; 97                     maze[i][j] = ‘.‘; 98                 } 99                 else if ( maze[i][j] == ‘^‘ )100                 {101                     ex = i;102                     ey = j;103                     maze[i][j] = ‘.‘;104                 }105             }106         }107         int ans = bfs();108         printf("%d\n", ans);109     }110     return 0;    111 }

 

hdu 1429 bfs+狀態壓縮