Lawnmower（除草）

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 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:

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)

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

划重点：他每走一行会换方向

#include <iostream>
#include <algorithm>
#include <cstring>using namespace std ;
//奇数行向右走
//偶数行向左走
int main()
{int n , m ; char a[160][160] ;cin >> n >> m ;for(int i = 0 ; i < n ; i ++)for(int j = 0 ; j < m ; j ++)cin >> a[i][j] ;int x = 0 , y = 0 ;//记录每一行的停留点int ans = 0 ;int j ;for(int i = 0 ; i < n; i ++)//每一行的，记录最后一株草的位置{if(i%2==0)j = 0 ;elsej = m-1 ; while(j >= 0 && j < m){if(a[i][j]=='W'){ans += abs(x-i)+abs(y-j) ;x = i , y = j ;}    if(i%2==0)j ++ ;elsej -- ;}}cout << ans << endl ;return 0 ;
}

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