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

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;

constexpr size_t maxn = 110;

int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};

int dl[][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int dr[][2] = {{0, 1}, {-1, 0 }, {0, -1}, {1, 0}};

int sx, sy, ex, ey, n, m;

char G[maxn][maxn];
struct Pos{
    int x, y, s;
};

int dfs(int x, int y, int d, int step, int dir[][2]){
    for(int i = 0; i < 4; ++ i){
        int j = ((d - 1 + 4)% 4 + i) % 4;
        int nx = x + dir[j][0];
        int ny = y + dir[j][1];

        if(nx == ex && ny == ey)return step+1;
        if(nx < 0 || ny < 0 || nx > n || ny > m)continue;
        if(G[nx][ny] == '#') continue;

        return dfs(nx, ny, j, step + 1, dir);
    }
}

int BFS(int sx, int sy){
	bool vis[maxn][maxn];
    memset(vis, false, sizeof(vis));

    queue<Pos> Q;
    Q.push({sx, sy, 1});
    vis[sx][sy] = true;

    while(!Q.empty()){
        Pos p = Q.front(); Q.pop();
        if(p.x == ex && p.y == ey) return p.s;
        Pos np;
        for(int i = 0; i < 4; ++ i){
            np.x = p.x + dx[i];
            np.y = p.y + dy[i];
            np.s = p.s + 1;
            if(np.x < 0 || np.x > n || np.y > m || np.y < 0)continue;
            if(vis[np.x][np.y])continue;
            if(G[np.x][np.y] != '#'){
                vis[np.x][np.y] = true;
                Q.push(np);
            }
        }
    }
}


int main(){
    int d1, d2;
    cin >> m >> n;
    for (int i = 0; i < n; ++ i){
        scanf("%s",  G[i]);
        for(int j = 0; j < m; ++ j){
            if(G[i][j] == 'S')sx = i, sy = j;
            else if(G[i][j] == 'E') ex = i, ey = j;
        }
    }
    if(sx == 0)d1 = 3, d2 = 3;
    else if(sx == n-1) d1 = 1, d2 = 1;
    else if(sy == 0)d1 = 2, d2 = 0;
    else if(sy == m-1)d1 = 0, d2 = 2;
    cout << dfs(sx, sy, d1, 1, dl) << endl;
    cout << dfs(sx, sy, d2, 1, dr) << endl;
    cout << BFS(sx, sy) << endl;
    return 0;

}