Codeforces 115 B Lawnmower【思维】

B. Lawnmower

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a garden consisting entirely of grass and weeds. Your garden is described by an n × m grid, with rows numbered 1 to n from top to bottom, and columns 1 to m from left to right. Each cell is identified by a pair (r, c) which means that the cell is located at row r and column c. Each cell may contain either grass or weeds. For example, a 4 × 5 garden may look as follows (empty cells denote grass):

You have a land-mower with you to mow all the weeds. Initially, you are standing with your lawnmower at the top-left corner of the garden. That is, at cell (1, 1). At any moment of time you are facing a certain direction — either left or right. And initially, you face right.

In one move you can do either one of these:

1) Move one cell in the direction that you are facing.

• if you are facing right: move from cell (r, c) to cell (r, c + 1)
  

• if you are facing left: move from cell (r, c) to cell (r, c - 1)
  

2) Move one cell down (that is, from cell (r, c) to cell (r + 1, c)), and change your direction to the opposite one.

• if you were facing right previously, you will face left
  

• if you were facing left previously, you will face right
  

You are not allowed to leave the garden. Weeds will be mowed if you and your lawnmower are standing at the cell containing the weeds (your direction doesn't matter). This action isn't counted as a move.

What is the minimum number of moves required to mow all the weeds?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 150) — the number of rows and columns respectively. Then follow n lines containing m characters each — the content of the grid. "G" means that this cell contains grass. "W" means that this cell contains weeds.

It is guaranteed that the top-left corner of the grid will contain grass.

Output

Print a single number — the minimum number of moves required to mow all the weeds.

Examples

Input

4 5
GWGGW
GGWGG
GWGGG
WGGGG

Output

11

Input

3 3
GWW
WWW
WWG

Output

7

Input

1 1
G

Output

0

Note

For the first example, this is the picture of the initial state of the grid:

A possible solution is by mowing the weeds as illustrated below:

题目大意：

按照题干中的四种方式走，只要向下走一行，方向就会转变，问最少走的步数能够除掉所有的草。

思路：

1、对应初始的时候是向右走的，首先需要向右走走到最右边的那个W的位子。

2、然后继续判断下一行，如果下一行有W在当前位子右边，还要推着小车向右多走几步。

3、对应同理，如果在向左走的时候，在当前行首先我们走到最左边的那个W的位子，然后再判断下一行，如果下一行有W在当前位子的左边，那么我们还要推着小车向左多走几步。

4、注意当割完所有草之后要跳出，要不然会有多余的行走步数。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[165][165];
int n,m;
int abs(int aaa)
{
    if(aaa<0)return -aaa;
    else return aaa;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%s",a[i]);
            for(int j=0; j<m; j++)
            {
                if(a[i][j]=='W')sum++;
            }
        }
        int d=1;
        int tmp=0;
        int output=0;
        int y=0;
        for(int i=0; i<n; i++)
        {
            int maxn=-1;
            //printf("%d %d %d\n",y,tmp,output);
            if(tmp==sum)break;
            for(int j=0; j<m; j++)
            {
                if(a[i][j]=='W')
                {
                    if(d==1)maxn=j;
                    else
                    {
                        if(maxn==-1)maxn=j;
                    }
                }
                if(a[i+1][j]=='W')
                {
                    if(d==1)maxn=j;
                    else
                    {
                        if(maxn==-1)maxn=j;
                    }
                }
            }
            if(maxn==-1)
            {
                output+=1;
                d=1-d;
                continue;
            }
            else
            {

                if(d==1)
                {
                    d=0;
                    output+=abs(y-maxn);
                    for(int z=0; z<m; z++)
                    {
                        if(a[i][z]=='W')tmp++;
                    }
                    y=maxn;
                    if(i!=n-1&&tmp!=sum)output++;
                }
                else
                {
                    d=1;
                    output+=abs(y-maxn);
                    for(int z=0; z<m; z++)
                    {
                        if(a[i][z]=='W')tmp++;
                    }
                    y=maxn;
                    if(i!=n-1&&tmp!=sum)output++;
                }
            }
        }
        printf("%d\n",output);
    }
}