题目描述

Farmer John has always done his best to keep the pastures full of luscious, delicious healthy grass for the cows. He has lost the battle, though, as the evil milkweed has attained a foothold in the northwest part of his farm.

The pasture, as usual, is partitioned into a rectilinear grid of height Y (1 <= Y <= 100) and width X (1 <= X <= 100) with (1,1) being in the lower left corner (i.e., arranged as a normal X,Y coordinate grid). The milkweed has initially begun growing at square (Mx,My). Each week the milkweed propagates to all non-rocky squares that surround any square it already occupies, as many as eight more squares (both the rectilinear squares and the diagonals). After only one week in those squares, it is ready to move on to more squares.

Bessie wants to enjoy all the grass she can before the pastures are taken over by milkweed. She wonders how long it can last. If the milkweed is in square (Mx,My) at time zero, at what time does it complete its invasion of the pasture (which, for the given input data, will always happen)?

The pasture is described by a picture with '.'s for grass and '*'s for boulders, like this example with X=4 and Y=3:

.... ..*.

.**. If the milkweed started in the lower left corner (row=1, column=1), then the map would progress like this:

.... .... MMM. MMMM MMMM

..*. MM*. MM*. MM*M MM*M

M. M. M. M. M**M

week 0 1 2 3 4

The milkweed has taken over the entire field after 4 weeks.

POINTS: 125

Farmer John一直努力让他的草地充满鲜美多汁的而又健康的牧草。可惜天不从人愿,他在植物大战人类中败下阵来。邪恶的乳草已经在他的农场的西北部份占领了一片立足之地。

草地像往常一样,被分割成一个高度为Y(1 <= Y <= 100), 宽度为X(1 <= X <= 100)的直角网格。(1,1)是左下角的格(也就是说坐标排布跟一般的X,Y坐标相同)。乳草一开始占领了格(Mx,My)。每个星期,乳草传播到已被乳草占领的格子四面八方的每一个没有很多石头的格(包括垂直与水平相邻的和对角在线相邻的格)。1周之后,这些新占领的格又可以把乳草传播到更多的格里面了。

Bessie想要在草地被乳草完全占领之前尽可能的享用所有的牧草。她很好奇到底乳草要多久才能占领整个草地。如果乳草在0时刻处于格(Mx,My),那么会在哪个时刻它们可以完全占领入侵整片草地呢?对给定的数据总是会发生。

输入输出格式

输入格式:

  • Line 1: Four space-separated integers: X, Y, Mx, and My

  • Lines 2..Y+1: Line y+1 describes row (Y+2-y) of the field with X characters ('.' for grass and '*' for a boulder)

输出格式:

  • Line 1: A single integer that is the week number when the milkweed takes over the last remaining non-boulder square of the pasture.

输入输出样例

输入样例#1: 复制

4 3 1 1
....
..*.
.**.
输出样例#1: 复制

4

思路:搜索。

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,sx,sy,ans;
char s[];
int vis[][];
int map[][];
int dx[]={-,-,-,,,,,};
int dy[]={-,,,-,,-,,};
struct nond{
int x,y,pos;
}v;
queue<nond>que;
int main(){
scanf("%d%d%d%d",&m,&n,&sy,&sx);
for(int i=n;i>=;i--){
scanf("%s",s);
for(int j=;j<=m;j++)
if(s[j-]=='*') map[i][j]=;
}
vis[sx][sy]=;nond tmp;
tmp.x=sx,tmp.y=sy;tmp.pos=;
que.push(tmp);
while(!que.empty()){
nond now=que.front();
que.pop();
for(int i=;i<;i++){
nond c;
c.x=now.x+dx[i];
c.y=now.y+dy[i];
c.pos=now.pos+;
if(c.x>=&&c.x<=n&&c.y>=&&c.y<=m&&!map[c.x][c.y]&&!vis[c.x][c.y]){
vis[c.x][c.y]=c.pos;
ans=max(ans,vis[c.x][c.y]);
que.push(c);
}
}
}
cout<<ans;
}

洛谷 P2960 [USACO09OCT]Milkweed的入侵Invasion of the Milkweed的更多相关文章

  1. 题解 P2960 【[USACO09OCT]Milkweed的入侵Invasion of the Milkweed】

    题目链接 首先这道题是一道经典的BFS.非常适合刚刚学习深搜的同学. 现在分析一下这个问题.首先,每周是八个方向.就是一圈. 也就是说入侵的范围关于时间是成辐射型扩散.让求最大时间. 也就是完美的BF ...

  2. 洛谷 P2376 [USACO09OCT]津贴Allowance 解题报告

    P2376 [USACO09OCT]津贴Allowance 题目描述 作为创造产奶纪录的回报,\(Farmer\) \(John\)决定开始每个星期给\(Bessie\)一点零花钱. \(FJ\)有一 ...

  3. 洛谷——P2958 [USACO09OCT]木瓜的丛林Papaya Jungle

    P2958 [USACO09OCT]木瓜的丛林Papaya Jungle 题目描述 Bessie has wandered off the farm into the adjoining farmer ...

  4. 洛谷—— P1339 [USACO09OCT]热浪Heat Wave

    P1339 [USACO09OCT]热浪Heat Wave 题目描述 The good folks in Texas are having a heatwave this summer. Their ...

  5. 洛谷 P2959 [USACO09OCT]悠闲漫步The Leisurely Stroll

    P2959 [USACO09OCT]悠闲漫步The Leisurely Stroll 题目描述 Bessie looks out the barn door at the beautiful spri ...

  6. 洛谷 P2958 [USACO09OCT]木瓜的丛林Papaya Jungle

    P2958 [USACO09OCT]木瓜的丛林Papaya Jungle 题目描述 Bessie has wandered off the farm into the adjoining farmer ...

  7. 洛谷P1339 [USACO09OCT]热浪Heat Wave 题解

    题目传送门 这道题实际非常简单好奇是怎么变黄的... 其实也就是一个SPFA,本人非常懒,不想打邻接表,直接用矩阵就好啦... #include<bits/stdc++.h> using ...

  8. 洛谷 P2639 [USACO09OCT]Bessie的体重问题Bessie's We… 题解

    题目传送门 这也是个01背包,只是装的很... #include<bits/stdc++.h> #define MAXN 45010 using namespace std; int f[ ...

  9. 洛谷 2957 [USACO09OCT]谷仓里的回声Barn Echoes

    题目描述 The cows enjoy mooing at the barn because their moos echo back, although sometimes not complete ...

随机推荐

  1. 新手须知 QT类大全

    QT类大全,在行内容中罗列出来了,希望大家多看看,如果是API就更好了,但可惜不是.这些是一些大类,请多做参考. QApplication 应用程序类 QLabel 标签类 QPushButton 按 ...

  2. 《一个民企CEO的职场阳谋》–读书总结(下)

    职场是一个战场,很多人几十年在这里战斗. 职场是一个熔炉,很多人大半生在这里修炼. 如果在办公室里得不到快乐,生活就不会快乐. 如果公司里头感觉不到幸福,人生就不会幸福.(以上四句来自老刘的博客) & ...

  3. Java线程:CountDownLatch 与Thread 的 join()

    需求: 主程序中需要等待所有子线程完成后 再继续任务 两种实现方式: 一种使用join() 方法:当在当前线程中调用某个线程 thread 的 join() 方法时,当前线程就会阻塞,直到thread ...

  4. Mysql学习总结(35)——Mysql两千万数据优化及迁移

    最近有一张2000W条记录的数据表需要优化和迁移.2000W数据对于MySQL来说很尴尬,因为合理的创建索引速度还是挺快的,再怎么优化速度也得不到多大提升.不过这些数据有大量的冗余字段和错误信息,极不 ...

  5. 【转】Visual Studio單元測試小應用-測執行時間

    [转]Visual Studio單元測試小應用-測執行時間 Visual Studio的單元測試會記錄每一個測試的執行時間,如果有幾個Method要測效能,以前我會用Stopwatch,最近我都改用單 ...

  6. 1.RunLoop是什么?

    1.Run loops是线程相关的的基础框架的一部分. 一个run loop就是一个事件处理的循环.用来不停的调度工作以及处理输入事件.使用run loop的目的是让你的线程在有工作的时候忙于工作.而 ...

  7. org.gradle.process.internal.ExecException: Process 'command 'C:\Program Files (x86)\Java\jdk1.7.0_7

    32为的androidstudio: build.gradle: dexOptions { javaMaxHeapSize "1g"}

  8. ThinkPHP5.0框架开发--第1章 Tp5.0安装

    ThinkPHP5.0框架开发--第1章 Tp5.0安装 第1章 Tp5.0 安装 ======================================================== 今 ...

  9. vue 初始化项目模板报错

    E:\xiaogezi.cn\vue>vue init webpack myProject vue-cli · Failed to download repo vuejs-templates/w ...

  10. C#调用webservice(一)

    最近一直在搞网络编程,这篇是关于webservice的,准备写两篇例子这篇是实现手机号码归宿地查询,下篇准备写实现机票查询. 这个网站(http://www.webxml.com.cn/zh_cn/i ...