[BFS][NOI题库][poj2157]Maze
Maze
| Time Limit: 2000MS | Memory Limit: 65536K |
Description
Acm, a treasure-explorer, is exploring again. This time he is in a special maze, in which there are some doors (at most 5 doors, represented by 'A', 'B', 'C', 'D', 'E' respectively). In order to find the treasure, Acm may need to open doors. However, to open a door he needs to find all the door's keys (at least one) in the maze first. For example, if there are 3 keys of Door A, to open the door he should find all the 3 keys first (that's three 'a's which denote the keys of 'A' in the maze). Now make a program to tell Acm whether he can find the treasure or not. Notice that Acm can only go up, down, left and right in the maze.
Input
The input consists of multiple test cases. The first line of each test case contains two integers M and N (1 < N, M < 20), which denote the size of the maze. The next M lines give the maze layout, with each line containing N characters. A character is one of the following: 'X' (a block of wall, which the explorer cannot enter), '.' (an empty block), 'S' (the start point of Acm), 'G' (the position of treasure), 'A', 'B', 'C', 'D', 'E' (the doors), 'a', 'b', 'c', 'd', 'e' (the keys of the doors). The input is terminated with two 0's. This test case should not be processed.
Output
For each test case, in one line output "YES" if Acm can find the treasure, or "NO" otherwise.
Sample Input
4 4 S.X. a.X. ..XG .... 3 4 S.Xa .aXB b.AG 0 0
Sample Output
YES NO
Source
POJ Monthly,Wang Yijie
大致题意：给定一张M行N列的图，探险家从起点S出发，探索财宝G。'A' 'B' 'C' 'D' 'E' 为带锁的门,需要收集到整张图上所有对应的钥匙'a' 'b' 'c' 'd' 'e' 才能打开门（例如，A门有3把钥匙在图上，如果要打开A门，他必须先找到所有三把属于A门的钥匙'a'）。问探险家能否找到宝藏。
分析：从S点出发BFS 更新手上的钥匙，与对应门的钥匙总量进行对比，如果相同即门被打开，将门加入队列。重复进行BFS直到到达G。
代码
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 #define MAX 25 5 using namespace std; 6 7 struct dui{ 8 int r, c; // hang lie 9 }queue[ MAX * MAX ]; 10 11 struct men{ 12 int key_sum, key_num; 13 }door[ 6 ]; 14 15 int m, n; 16 char map[ MAX ][ MAX ]; 17 bool vis[ MAX ][ MAX ]; 18 bool vis_door[ MAX ][ MAX ]; 19 20 int wx[ 5 ] = { 0, -1, 0, 1, 0}; 21 int wy[ 5 ] = { 0 ,0, 1, 0, -1}; 22 23 bool able( int x, int y ) 24 { 25 if( x > 0 && x <= m && y > 0 && y <= n) 26 return true; 27 else return false; 28 } 29 30 int main() 31 { 32 int i, j; 33 int r, c; 34 char tmp, tt; 35 while( true ){ 36 scanf("%d%d", &m ,&n); 37 if( m == 0 && n == 0 ) break; 38 memset( vis, 0, sizeof( vis )); 39 memset( door, 0, sizeof( door )); 40 memset( vis_door, 0, sizeof( vis_door )); 41 for( i = 1; i <= m; i++ ){ 42 scanf("%s", map[ i ] + 1); 43 for( j = 1; j <= n; j++ ){ 44 if( map[ i ][ j ] >= 'a' && map[ i ][ j ] <= 'e'){ 45 door[ map[ i ][ j ] - 'a' + 1 ].key_sum ++; 46 } 47 if( map[ i ][ j ] == 'S' ){ 48 queue[ 0 ].r = i; 49 queue[ 0 ].c = j; 50 vis[ i ][ j ] = 1; 51 } 52 } 53 } 54 55 int head = 0, tail = 1; 56 bool finded = false; 57 bool newdoor = true; 58 59 while( newdoor ){ 60 while( !finded && head < tail){ 61 for( i = 1; i <= 4; i++ ){ 62 r = queue[ head ].r + wx[ i ]; 63 c = queue[ head ].c + wy[ i ]; 64 65 if( map[ r ][ c ] == 'G' ){ 66 finded = true; 67 break; 68 } 69 70 if( !vis[ r ][ c ] && able( r, c ) && map[ r ][ c ] != 'X'){ 71 tmp = map[ r ][ c ]; 72 73 if( tmp >= 'A' && tmp <= 'E'){ 74 vis_door[ r ][ c ] = true; 75 continue; 76 } 77 78 if( tmp >= 'a' && tmp <= 'e' ){ 79 door[ tmp - 'a' + 1 ].key_num++; 80 } 81 vis[ r ][ c ] = true; 82 queue[ tail ].r = r; 83 queue[ tail ].c = c; 84 tail++; 85 } 86 } 87 head++; 88 } 89 if( finded ) break; 90 91 newdoor = false; 92 93 for( r = 1; r <= m; r++ ){ 94 for( c = 1; c <= n; c++ ){ 95 tmp = map[ r ][ c ]; 96 if( tmp >= 'A' && tmp <= 'E' && vis_door[ r ][ c ] && !vis[ r ][ c ]){ 97 tt = tmp - 'A' + 1; 98 if( door[ tt ].key_sum > 0 && door[ tt ].key_sum == door[ tt ].key_num ){ 99 vis[ r ][ c ] = true; 100 queue[ tail ].r = r; 101 queue[ tail ].c = c; 102 tail++; 103 newdoor = true; 104 } 105 } 106 } 107 } 108 } 109 110 if( finded ) printf("YES\n"); 111 else printf("NO\n"); 112 } 113 114 return 0; 115 }
