问题可以抽象为求起点到终点的最短路，其中经过x的代价是0，而经过.的代价是1，bfs的时候用优先队列即可。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <queue> 5 using namespace std; 6 7 const int N = 1001; 8 char maze[N][N]; 9 bool visit[N][N]; 10 int dx[] = { 0, 0, 1, -1 }; 11 int dy[] = { 1, -1, 0, 0 }; 12 int n, m, sx, sy, ex, ey; 13 14 bool ok( int x, int y ) 15 { 16 return x >= 0 && x < n && y >= 0 && y < m && !visit[x][y]; 17 } 18 19 struct Node 20 { 21 int x, y, t; 22 Node(){} 23 Node( int _x, int _y, int _t ) 24 { 25 x = _x, y = _y, t = _t; 26 } 27 bool operator < ( const Node & o ) const 28 { 29 return t > o.t; 30 } 31 }; 32 33 priority_queue<Node> q; 34 35 int bfs() 36 { 37 while ( !q.empty() ) q.pop(); 38 memset( visit, 0, sizeof(visit) ); 39 q.push( Node( sx, sy, 0 ) ); 40 visit[sx][sy] = 1; 41 while ( !q.empty() ) 42 { 43 Node cur = q.top(); 44 q.pop(); 45 if ( cur.x == ex && cur.y == ey ) return cur.t; 46 for ( int i = 0; i < 4; i++ ) 47 { 48 int x = cur.x + dx[i]; 49 int y = cur.y + dy[i]; 50 if ( ok( x, y ) ) 51 { 52 int d = maze[x][y] == '.' ? 1 : 0; 53 q.push( Node( x, y, cur.t + d ) ); 54 visit[x][y] = 1; 55 } 56 } 57 } 58 return -1; 59 } 60 61 int main () 62 { 63 while ( scanf("%d%d", &n, &m) != EOF ) 64 { 65 if ( !n && !m ) break; 66 for ( int i = 0; i < n; i++ ) 67 { 68 scanf("%s", maze[i]); 69 } 70 scanf("%d%d%d%d", &sx, &sy, &ex, &ey); 71 sx--, sy--, ex--, ey--; 72 int ans = bfs(); 73 printf("%d ", ans); 74 } 75 return 0; 76 }