Codeforces 128 A Statues【预处理+Bfs】

A. Statues

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In this task Anna and Maria play a game with a very unpleasant rival. Anna and Maria are in the opposite squares of a chessboard (8 × 8): Anna is in the upper right corner, and Maria is in the lower left one. Apart from them, the board has several statues. Each statue occupies exactly one square. A square that contains a statue cannot have anything or anyone — neither any other statues, nor Anna, nor Maria.

Anna is present on the board as a figurant (she stands still and never moves), and Maria has been actively involved in the game. Her goal is — to come to Anna's square. Maria and statues move in turn, Maria moves first. During one move Maria can go to any adjacent on the side or diagonal cell in which there is no statue, or she can stay in the cell where she is. The statues during their move must go one square down simultaneously, and those statues that were in the bottom row fall from the board and are no longer appeared.

At that moment, when one of the statues is in the cell in which the Maria is, the statues are declared winners. At the moment when Maria comes into the cell where Anna has been waiting, Maria is declared the winner.

Obviously, nothing depends on the statues, so it all depends on Maria. Determine who will win, if Maria does not make a strategic error.

Input

You are given the 8 strings whose length equals 8, describing the initial position on the board. The first line represents the top row of the board, the next one — for the second from the top, and so on, the last line represents the bottom row. Each character string matches a single cell board in the appropriate row, and the characters are in the same manner as that of the corresponding cell. If the cell is empty, the corresponding character is ".". If a cell has Maria, then it is represented by character "M". If a cell has Anna, it is represented by the character "A". If a cell has a statue, then the cell is represented by character "S".

It is guaranteed that the last character of the first row is always "A", the first character of the last line is always "M". The remaining characters are "." or "S".

Output

If Maria wins, print string "WIN". If the statues win, print string "LOSE".

Examples

Input

.......A................................................M.......

Output

WIN

Input

.......A........................................SS......M.......

Output

LOSE

Input

.......A.................................S......S.......MS......

Output

LOSE

题目大意：

主角开局位于M处，他如果能够走到A处就是WIN，否则就是Lose。

主角先走一步之后， 所有的S下落一层，M不能走S，而且每轮有9种走法：向周围八个地方移动，以及原地不动。

思路：

1、首先预处理出所有S下落的时间，对应设定一个vector<int >mp【i】【j】，表示点（i，j）处会出现落石的时间集合。

2、然后我们Bfs起点，一点一点向A走，如果能够走到，输出WIN，否则输出LOSE即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;struct node{    int x,y,step;}now,nex;char a[10][10];int vis[10][10];vector<int >mp[10][10];int fx[9]={0,0,1,-1,0,1,1,-1,-1};int fy[9]={1,-1,0,0,0,1,-1,1,-1};int n=8,m=8;void Bfs(int sx,int sy){    queue<node >s;    now.x=sx;    now.y=sy;    s.push(now);    memset(vis,0,sizeof(vis));    vis[sx][sy]=1;    while(!s.empty())    {        now=s.front();        if(a[now.x][now.y]=='A')        {            printf("WIN\n");            return ;        }        s.pop();        for(int i=0;i<9;i++)        {            nex.x=now.x+fx[i];            nex.y=now.y+fy[i];            nex.step=now.step;            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m)            {                if(i!=4&&vis[nex.x][nex.y]==1)continue;                int flag=0;                for(int k=0;k<mp[nex.x][nex.y].size();k++)                {                    if(nex.step==mp[nex.x][nex.y][k]||nex.step+1==mp[nex.x][nex.y][k])flag=1;                }                if(flag==0)                {                    nex.step+=1;                    s.push(nex);                    vis[nex.x][nex.y]=1;                }            }        }    }    printf("LOSE\n");    return ;}int main(){    for(int i=0;i<8;i++)    {        scanf("%s",a[i]);    }    int sx,sy;    for(int i=0;i<8;i++)    {        for(int j=0;j<8;j++)        {            if(a[i][j]=='M')            {                sx=i;sy=j;            }            if(a[i][j]=='S')            {                mp[i][j].push_back(0);                for(int k=1;k<8;k++)                {                    int xx=i+k;                    int y=j;                    if(xx<8)                    {                        mp[xx][y].push_back(k);                    }                }            }        }    }    Bfs(sx,sy);}