BFS+状态压缩，做了很多状态压缩了。今晚把八数码问题给搞定了。
1 #include <iostream> 2 #include <queue> 3 #include <cstring> 4 #include <cstdio> 5 using namespace std; 6 7 typedef struct node_st { 8 int x, y, t; 9 int key; 10 node_st() {} 11 node_st(int xx, int yy, int tt, int kk) { 12 x = xx; y = yy; t = tt; key = kk; 13 } 14 } node_st; 15 16 char map[25][25]; 17 char visit[1<<10][25][25]; 18 int direct[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 19 int n, m, time; 20 21 int bfs(int bx, int by) { 22 int x, y, key, t; 23 int i, tmp; 24 queue<node_st> que; 25 node_st node; 26 27 memset(visit, 0, sizeof(visit)); 28 visit[bx][by][0] = 1; 29 que.push(node_st(bx,by,0,0)); 30 31 while ( !que.empty() ) { 32 node = que.front(); 33 if (node.t >= time) 34 return -1; 35 if (map[node.x][node.y] == '^') 36 return node.t; 37 que.pop(); 38 t = node.t + 1; 39 for (i=0; i<4; ++i) { 40 x = node.x + direct[i][0]; 41 y = node.y + direct[i][1]; 42 if (x<0 || x>=n || y<0 || y>=m) 43 continue; 44 if (map[x][y]>='a' && map[x][y]<='j') 45 key = node.key | (1<<(map[x][y]-'a')); 46 else 47 key = node.key; 48 if (visit[key][x][y] || map[x][y]=='*') 49 continue; 50 if (map[x][y]>='A' && map[x][y]<='J') { 51 tmp = (1<<(map[x][y]-'A')) & key; 52 if ( !tmp ) 53 continue; 54 } 55 que.push(node_st(x,y,t,key)); 56 visit[key][x][y] = 1; 57 } 58 } 59 60 return -1; 61 } 62 63 int main() { 64 int bx, by; 65 int i, j; 66 67 while (scanf("%d %d %d", &n, &m, &time) != EOF) { 68 for (i=0; i<n; ++i) { 69 scanf("%s", map[i]); 70 for (j=0; j<m; ++j) { 71 if (map[i][j] == '@') { 72 bx = i; 73 by = j; 74 } 75 } 76 } 77 i = bfs(bx, by); 78 printf("%d ", i); 79 } 80 81 return 0; 82 }