Lawnmower

B - Lawnmower

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 rand 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.

没看懂题之前觉得挺难的，明白题意之后发现其实很简单的一个题，就是奇数行的时候向右除草，偶数行的时候向左除草（但是因为我的坐标是从0开始的，所以计算时相反）当遇到一个杂草就用杂草的坐标减去前一个杂草的坐标，当换行时也要加一，去绝对值之和。

#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;


int main()
{
    int n,m;
    char a[155][155];
    while(~scanf("%d%d",&n,&m))
    {
        int i,j;
        for(i=0;i<n;i++)
            cin>>a[i];
        int ans=0,x=0,y=0;
        for(i=0;i<n;i++)
        {
            if(i%2==0){                                                         //奇数行
            for(j=0;j<m;j++)
            {
                if(a[i][j]=='W')
                {
                    ans+=fabs(i-x)+fabs(j-y);                         //向右除草
                    x=i;y=j;                                                       //记录下当前杂草的坐标
                }
            }
            }
            else{
            for(j=m-1;j>=0;j--)
            {
                if(a[i][j]=='W')
                {
                    ans+=fabs(i-x)+fabs(j-y);                         //向左除草
                    x=i;y=j;
                }
            }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}