题目盲人摸象【dfs,bfs, 顺序遍历】

此题就是方向遍历，始终是想遍历左手所指得位置，或右手所指得位置。

也就是一个遍历数组得顺序问题

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <queue>
  5 using namespace std;
  6 
  7 constexpr size_t maxn = 110;
  8 
  9 int dx[] = {0, -1, 0, 1};
 10 int dy[] = {-1, 0, 1, 0};
 11 
 12 int dl[][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
 13 int dr[][2] = {{0, 1}, {-1, 0 }, {0, -1}, {1, 0}};
 14 
 15 int sx, sy, ex, ey, n, m;
 16 
 17 char G[maxn][maxn];
 18 struct Pos{
 19     int x, y, s;
 20 };
 21 
 22 int dfs(int x, int y, int d, int step, int dir[][2]){
 23     for(int i = 0; i < 4; ++ i){
 24         int j = ((d - 1 + 4)% 4 + i) % 4;
 25         int nx = x + dir[j][0];
 26         int ny = y + dir[j][1];
 27 
 28         if(nx == ex && ny == ey)return step+1;
 29         if(nx < 0 || ny < 0 || nx > n || ny > m)continue;
 30         if(G[nx][ny] == '#') continue;
 31 
 32         return dfs(nx, ny, j, step + 1, dir);
 33     }
 34 }
 35 
 36 int BFS(int sx, int sy){
 37 	bool vis[maxn][maxn];
 38     memset(vis, false, sizeof(vis));
 39 
 40     queue<Pos> Q;
 41     Q.push({sx, sy, 1});
 42     vis[sx][sy] = true;
 43 
 44     while(!Q.empty()){
 45         Pos p = Q.front(); Q.pop();
 46         if(p.x == ex && p.y == ey) return p.s;
 47         Pos np;
 48         for(int i = 0; i < 4; ++ i){
 49             np.x = p.x + dx[i];
 50             np.y = p.y + dy[i];
 51             np.s = p.s + 1;
 52             if(np.x < 0 || np.x > n || np.y > m || np.y < 0)continue;
 53             if(vis[np.x][np.y])continue;
 54             if(G[np.x][np.y] != '#'){
 55                 vis[np.x][np.y] = true;
 56                 Q.push(np);
 57             }
 58         }
 59     }
 60 }
 61 
 62 
 63 int main(){
 64     int d1, d2;
 65     cin >> m >> n;
 66     for (int i = 0; i < n; ++ i){
 67         scanf("%s",  G[i]);
 68         for(int j = 0; j < m; ++ j){
 69             if(G[i][j] == 'S')sx = i, sy = j;
 70             else if(G[i][j] == 'E') ex = i, ey = j;
 71         }
 72     }
 73     if(sx == 0)d1 = 3, d2 = 3;
 74     else if(sx == n-1) d1 = 1, d2 = 1;
 75     else if(sy == 0)d1 = 2, d2 = 0;
 76     else if(sy == m-1)d1 = 0, d2 = 2;
 77     cout << dfs(sx, sy, d1, 1, dl) << endl;
 78     cout << dfs(sx, sy, d2, 1, dr) << endl;
 79     cout << BFS(sx, sy) << endl;
 80     return 0;
 81 
 82 }
 83