训练赛1_E_Lawnmower

探讨在一个由草地和杂草组成的网格中,如何使用最少的移动步骤完成割草任务。通过智能路径规划,避免重复行走,实现高效割草。

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):

img

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)

    img

  • if you are facing left: move from cell

    (r, c)

    to cell

    (r, c - 1)

    img

  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

    img

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

    img

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

Copy

4 5
GWGGW
GGWGG
GWGGG
WGGGG

output

Copy

11

input

Copy

3 3
GWW
WWW
WWG

output

Copy

7

input

Copy

1 1
G

output

Copy

0

Note

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

img

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

img

题目大意

一个人割草,输入一个n×mn \times mn×m的图,G代表空地,W代表草,奇数行只能往右走,偶数行只能往左走。问割完草的最少步数。

题目分析

之前各种想,各种特殊情况处理特判……其实只要知道一个矩形图,一个角走到另一个角,只要不往回走,其实走的路径就是一样的。不用考虑一定是往那条路走。那么只要模拟一下走的过程,记录一下步数就行了。我们从左上角走,然后往右遍历找到就计算一下距离,更新一下当前的位置,然后第二行从右往左扫。第三行从左往右扫,如此循环即可。

代码

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef struct{
  int x, y;
}node;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int n, m, x, y, k;
ll f(node v){
  int a = inf, b = inf;
  if(v.x < 0)
    a = (x - 1) / abs(v.x);
  else if(v.x > 0)
    a = (n - x) / abs(v.x);
  if(v.y < 0)
    b = (y - 1) / abs(v.y);
  else if(v.y > 0)
    b = (m - y) / abs(v.y);
  int mn = min(a, b);
  x += mn * v.x; y += mn * v.y;
  return mn;
}
int main(int argc, char const *argv[]) {
  scanf("%d%d%d%d", &n, &m, &x, &y);
  scanf("%d", &k);
  ll ans = 0;
  for(int i = 0; i < k; i++){
    node t;
    scanf("%d%d", &t.x, &t.y);
    ans += f(t);
  }
  printf("%I64d\n", ans);
  return 0;
}

基于可靠性评估序贯蒙特卡洛模拟法的配电网可靠性评估研究(Matlab代码实现)内容概要:本文围绕“基于可靠性评估序贯蒙特卡洛模拟法的配电网可靠性评估研究”,介绍了利用Matlab代码实现配电网可靠性的仿真分析方法。重点采用序贯蒙特卡洛模拟法对配电网进行长时间段的状态抽样与统计,通过模拟系统元件的故障与修复过程,评估配电网的关键可靠性指标,如系统停电频率、停电持续时间、负荷点可靠性等。该方法能够有效处理复杂网络结构与设备时序特性,提升评估精度,适用于含分布式电源、电动汽车等新型负荷接入的现代配电网。文中提供了完整的Matlab实现代码与案例分析,便于复现和扩展应用。; 适合人群:具备电力系统基础知识和Matlab编程能力的高校研究生、科研人员及电力行业技术人员,尤其适合从事配电网规划、运行与可靠性分析相关工作的人员; 使用场景及目标:①掌握序贯蒙特卡洛模拟法在电力系统可靠性评估中的基本原理与实现流程;②学习如何通过Matlab构建配电网仿真模型并进行状态转移模拟;③应用于含新能源接入的复杂配电网可靠性定量评估与优化设计; 阅读建议:建议结合文中提供的Matlab代码逐段调试运行,理解状态抽样、故障判断、修复逻辑及指标统计的具体实现方式,同时可扩展至不同网络结构或加入更多不确定性因素进行深化研究。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值