Codeforces 196 B. Infinite Maze（bfs）

野性酷女 2022-06-02 05:25 149阅读 0赞

We've got a rectangular *n* × *m*\-cell maze. Each cell is either passable, or is a wall (impassable). A little boy found the maze and cyclically tiled a plane with it so that the plane became an infinite maze. Now on this plane cell (*x*, *y*) is a wall if and only if cell ![22b1e94f2e7fa46812844b51d62d5961_v_1515033962][] is a wall.

In this problem ![eb50f2ebccba4d338cdc8f47f4cec6f8_v_1515033962][] is a remainder of dividing number *a* by number *b*.

The little boy stood at some cell on the plane and he wondered whether he can walk infinitely far away from his starting position. From cell (*x*, *y*) he can go to one of the following cells: (*x*, *y* - 1), (*x*, *y* + 1), (*x* - 1, *y*) and (*x* + 1, *y*), provided that the cell he goes to is not a wall.

Input

The first line contains two space-separated integers *n* and *m* (1 ≤ *n*, *m* ≤ 1500) — the height and the width of the maze that the boy used to cyclically tile the plane.

Each of the next *n* lines contains *m* characters — the description of the labyrinth. Each character is either a "\#", that marks a wall, a ".", that marks a passable cell, or an "S", that marks the little boy's starting point.

The starting point is a passable cell. It is guaranteed that character "S" occurs exactly once in the input.

Output

Print "Yes" (without the quotes), if the little boy can walk infinitely far from the starting point. Otherwise, print "No" (without the quotes).

Example

Input

5 4
    ##.#
    ##S#
    #..#
    #.##
    #..#

Output

Yes

Input

5 4
    ##.#
    ##S#
    #..#
    ..#.
    #.##

Output

Note

In the first sample the little boy can go up for infinitely long as there is a "clear path" that goes vertically. He just needs to repeat the following steps infinitely: up, up, left, up, up, right, up.

In the second sample the vertical path is blocked. The path to the left doesn't work, too — the next "copy" of the maze traps the boy.

题解：

这题的思路就是只要能从一张图的一个位置走到另一个相同位置但是坐标不一样就算可以到无限远了，一开始以为很简单的宽展个9张图就可以了。。然后到25组就wa了

后来仔细想想不止9张图，就用数组visx\[i\]\[j\]记录下经历这张图的这个位置x，visy\[i\]\[j\]记录位置y，只要是第二次经过一张图的同一个位置就输出yes

代码：

#include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<queue>
    #include<stack>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<stdlib.h>
    #include<cmath>
    #include<string>
    #include<algorithm>
    #include<iostream>
    #include<stdio.h>
    using namespace std;
    #define ll long long
    int n,m;
    struct node
    {
        int x,y;
    };
    int dirx[4]={0,0,-1,1};
    int diry[4]={-1,1,0,0};
    int visx[1505][1505];
    int visy[1505][1505];
    char s[1505][1505];
    queue<node>q;
    int main()
    {
        int i,j,k,stx,sty;
        node now,next;
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%s",s[i]);
            for(j=0;j<m;j++)
            {
                if(s[i][j]=='S')
                {
                    stx=i;
                    sty=j;
                }
                visx[i][j]=10086111;
                visy[i][j]=10086111;
            }
        }
        now.x=stx;
        now.y=sty;
        int tag=0;
        visx[stx][sty]=stx;
        visy[stx][sty]=sty;
        q.push(now);
        int xx,yy;
        while(!q.empty())
        {
            now=q.front();
            q.pop();
            for(i=0;i<4;i++)
            {
                next.x=now.x+dirx[i];
                next.y=now.y+diry[i];
                xx=(next.x%n+n)%n;
                yy=(next.y%m+m)%m;
                if(s[xx][yy]!='#')
                {
                    if(visx[xx][yy]==10086111)//表示之前没有经历过，随便设一个特殊值
                    {
                        visx[xx][yy]=next.x;
                        visy[xx][yy]=next.y;
                        q.push(next);
                    }
                    else if(visx[xx][yy]!=next.x||visy[xx][yy]!=next.y)
                    {
                        tag=1;
                        goto loop;
                    }
                }
            }
        }
        loop:;
        if(tag)
            printf("Yes\n");
        else
            printf("No\n");
        return 0;
    }

[22b1e94f2e7fa46812844b51d62d5961_v_1515033962]: https://odzkskevi.qnssl.com/22b1e94f2e7fa46812844b51d62d5961?v=1515033962
[eb50f2ebccba4d338cdc8f47f4cec6f8_v_1515033962]: https://odzkskevi.qnssl.com/eb50f2ebccba4d338cdc8f47f4cec6f8?v=1515033962