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#include
   
    #include
    
     #define MAX 22#include
     
      using namespace std;struct Node{	int x, y, s;	friend bool operator<(const Node &a, const Node &b){		return a.s > b.s;	}};int vis[MAX][MAX];char c;int dir[4][2] = {
  {1,0},{-1,0},{0,1},{0,-1}};int n, m;int sx, sy;char map[MAX][MAX];int check(int x, int y){	if(x>=0 && x
      
       =0 && y
       
         pq;	node.x = x; node.y = y; node.s = 0;	pq.push(node);	vis[node.x][node.y] = 1;//标记已经访问	while(!pq.empty()){		t1 = pq.top();		pq.pop();		if(map[t1.x][t1.y] == 'T') 			return t1.s;		//if(map[t2.x][t2.y] == 'T') return t2.s;		for(int i = 0; i < 4; ++i){			int nx = t1.x + dir[i][0];			int ny = t1.y + dir[i][1];			if(!check(nx,ny)) continue;			t2.x = nx; t2.y = ny; t2.s = t1.s+1;			//下一位置是楼梯时候才考虑 否则正常走 			//有楼梯时*********************************			if(map[nx][ny] == '|' || map[nx][ny] == '-'){				if(t2.s%2 == 0){ //这里为偶数步时候 其之前一步的楼梯应该反向  					if(map[nx][ny] == '|')						c = '-';					else if(map[nx][ny] == '-')						c = '|';					}				else					c = map[nx][ny];				t2.x += dir[i][0];				t2.y += dir[i][1];				if(!check(t2.x,t2.y)) continue;				if(c == '|' && (dir[i][1] == 1 || dir[i][1] == -1) || c == '-'&& (dir[i][0] == 1 || dir[i][0] == -1)){					t2.s += 1;//c表示到达楼梯前一步时候楼梯所处的状态							}			//如果人沿着x（上下）走 楼梯横着 等待							//人沿着y（左右）走 楼梯 竖着等待 			}		//***********************************	                vis[t2.x][t2.y] = 1;                pq.push(t2);		}		}//end OF while 	return -1;	}int main()	{		while(scanf("%d%d",&n,&m)!=EOF){			for(int i = 0; i < n; ++i){				for(int j = 0; j < m; ++j){					cin >> map[i][j];					if(map[i][j] == 'S'){ 						sx = i; sy = j;  					}				}			}			memset(vis,0,sizeof(vis));			int ans = bfs(sx, sy);			printf("%d\n",ans);		}					return 0;	}
       
      
     
    
   

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