6 6
..X...
XXX.X.
....X.
X.....
X.....
X.X...
3 5 6 3
如上一个图 告诉起点和终点 X到达不费力气 .到达花费1 问从起点到终点 最少的花费力气
分析：bfs 遇到X在bfs
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 7 const int maxn = 1005; 8 char s[maxn][maxn]; 9 int vis[maxn][maxn]; 10 int n, m; 11 int x0, y0, xn, yn; 12 13 struct Node { 14 int x, y, id; 15 bool operator<(const Node &n0) const{ 16 return id > n0.id; 17 } 18 }; 19 priority_queue<Node> q, qq; 20 21 int xx[4] = { 0, 0, 1, -1 }; 22 int yy[4] = { 1, -1, 0, 0 }; 23 24 bool bfs2(Node n0) { 25 while(!qq.empty()) { 26 qq.pop(); 27 } 28 qq.push(n0); 29 Node n2; 30 while(!qq.empty()) { 31 Node n1 = qq.top(); qq.pop(); 32 if(n1.x == xn && n1.y == yn) return true; 33 for(int i = 0; i < 4; i++) { 34 int tx = n1.x + xx[i]; 35 int ty = n1.y + yy[i]; 36 if(tx >= 0 && tx < n && ty >= 0 && ty < m && !vis[tx][ty] && s[tx][ty] == 'X') { 37 vis[tx][ty] = vis[n1.x][n1.y]; 38 n2.x = tx; n2.y = ty; n2.id = vis[tx][ty]; 39 qq.push(n2); 40 q.push(n2); 41 } 42 } 43 } 44 return false; 45 } 46 47 48 void bfs() { 49 while(!q.empty()) q.pop(); 50 memset(vis, 0, sizeof(vis)); 51 vis[x0][y0] = 1; 52 Node n0 = (Node) {x0, y0, vis[x0][y0]}; 53 q.push(n0); 54 if(s[x0][y0] == 'X') { 55 if(bfs2(n0)) return; 56 } 57 Node n2; 58 while(!q.empty()) { 59 Node n1 = q.top();q.pop(); 60 if(n1.x == xn && n1.y == yn) return ; 61 for(int i = 0; i < 4; i++) { 62 int tx = n1.x + xx[i]; 63 int ty = n1.y + yy[i]; 64 if(tx >= 0 && tx < n && ty >= 0 && ty < m && !vis[tx][ty]) { 65 if(s[tx][ty] == 'X') { 66 vis[tx][ty] = vis[n1.x][n1.y]; 67 n2.x = tx; n2.y = ty; n2.id = vis[tx][ty]; 68 if(bfs2(n2)) return ; 69 q.push(n2); 70 } else { 71 vis[tx][ty] = vis[n1.x][n1.y] + 1; 72 n2.x = tx; n2.y = ty; n2.id = vis[tx][ty]; 73 q.push(n2); 74 } 75 } 76 } 77 } 78 } 79 80 81 int main() { 82 while(scanf("%d %d",&n, &m) ) { 83 if(n == 0 && m == 0) break; 84 for(int i = 0; i < n; i++) { 85 scanf("%s",s[i]); 86 } 87 scanf("%d %d %d %d",&x0, &y0, &xn, &yn); 88 x0--; y0--; xn--; yn--; 89 bfs(); 90 printf("%d ", vis[xn][yn] - 1); 91 } 92 return 0; 93 }