B. Igor and his way to work

题目链接

题意

就是给你一个1000 * 1000以内的迷宫，只能转弯两次是否能够从起点到终点。

解题思路

题意还是十分清晰的。以前没做过转弯的题目，所以花了比较久的时间。题目不能。在原来bfs的基础上修改判断
重复遍历的条件。改为转弯次数是否小于前一次遍历的次数。如果大于便不用进队列了。因为上次在已经转弯的次
数小于这次的情况下都没有走通，那么这次次数少的情况下走依然不会走通。

代码

    #include <iostream>
    #include <stdio.h>
    #include <stack>
    #include <queue>
    #include <string.h>
    using namespace std;

    const int maxn = 1005;
    int stepr[4] = {1,0,-1,0};
    int stepl[4] = {0,1,0,-1};
    char mymap[maxn][maxn];
    int visit[maxn][maxn] = {0};
    struct node
    {
        int x,y,times;
        int type;
    };
    queue<node> s;
    bool solve(int m, int n)
    {
        memset(visit, 3,sizeof(visit));
        int sx,sy,ex,ey;
        for(int i = 0; i < m; i ++)
            for(int j = 0; j < n; j++)
            {
                if(mymap[i][j] == 'S')
                {
                    sx = i;
                    sy = j;
                }
                if(mymap[i][j] == 'T')
                {
                    ex = i;
                    ey = j;
                }
            }
        node temp,next;
        temp.times = 0;
        temp.x = sx;
        temp.y = sy;
        temp.type = 4;
        s.push(temp);
        int tx,ty;
        while(!s.empty())
        {
            temp = s.front();
            s.pop();
            if(temp.times > 2)
                continue;
            for(int i = 0; i < 4; i++)
            {
                tx = temp.x + stepr[i];
                ty = temp.y + stepl[i];
                if(tx >= 0 && tx < m && ty >= 0 && ty < n && mymap[tx][ty] != '*' && visit[tx][ty] > temp.times)
                {

                    //visit[tx][ty] = 1;
                    next.type = i;
                    if(i != temp.type && temp.type != 4)
                        next.times = temp.times+1;
                    else next.times = temp.times;
                    next.x = tx;
                    next.y = ty;
                    if(next.times > 2);
                    else s.push(next);

                    if(next.times > 2)
                        continue;
                    if(tx == ex && ty == ey)
                        return true;
                    visit[tx][ty] = next.times;
                }
            }
        }

        return false;
    }
    int main()
    {
        int m,n;
        scanf("%d %d", &m,&n);
        for(int i = 0; i < m; i++)
        {
            scanf("%s",mymap[i]);
        }
        if(solve(m,n))
        {
            cout << "YES" <<endl;
        }
        else cout << "NO" << endl;
        return 0;
    }