Codeforces Gym 100002 C "Cricket Field" 暴力
"Cricket Field"
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100002
Description
Once upon a time there was a greedy King who ordered his chief Architect to build a field for royal cricket inside his park. The King was so greedy, that he would not listen to his Architect's proposals to build a field right in the park center with pleasant patterns of trees specially planted around and beautiful walks inside tree alleys for spectators. Instead, he ordered neither to cut nor to plant even a single tree in his park, but demanded to build the largest possible cricket field for his pleasure. If the Kind finds that the Architect has dared to touch even a single tree in his park or designed a smaller field that it was possible, then the Architect will loose his head. Moreover, he demanded his Architect to introduce at once a plan of the field with its exact location and size.
Your task is to help poor Architect to save his head, by writing a program that will find the maximum possible size of the cricket field and its location inside the park to satisfy King's requirements.
The task is somewhat simplified by the fact, that King's park has a rectangular shape and is situated on a flat ground. Moreover, park's borders are perfectly aligned with North-South and East-West lines. At the same time, royal cricket is always played on a square field that is also aligned with North-South and East-West lines. Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of every tree. This coordinate system is, of course, aligned with North-South and East-West lines. Southwestern corner of the park has coordinates (0, 0) and Northeastern corner of the part has coordinates (W, H), where W and H are the park width and height in feet respectively.
For this task, you may neglect the diameter of the trees. Trees cannot be inside the cricket field, but may be situated on its side. The cricket field may also touch park's border, but shall not lie outside the park.
Input
The first line of the input file contains three integer numbers N, W, and H, separated by spaces. N (0 ≤ N ≤ 100) is the number of trees in the park. W and H (1 ≤ W, H ≤ 10000) are the park width and height in feet respectively.
Next N lines describe coordinates of trees in the park. Each line contains two integer numbers Xi and Yi separated by a space (0 ≤ Xi ≤ W, 0 ≤ Yi ≤ H) that represent coordinates of ith tree. All trees are located at different coordinates.
Output
Write to the output file a single line with three integer numbers P, Q, and L separated by spaces, where (P, Q) are coordinates of the cricket field Southwestern corner, and L is a length of its sides. If there are multiple possible field locations with a maximum size, then output any one.
Sample Input
7 10 7
3 2
4 2
7 0
7 3
4 5
2 4
1 7
Sample Output
4 3 4
HINT
题意
给你个nm的区域,问你最大的内部没有树的正方形有多大(边界上可以有树
题解:
读入Y轴之后,暴力枚举Y轴
然后统计在这个区域的正方形最大是多少
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1) using namespace std;
const int maxn = 1e2 + ;
int n , w , h ,ansx = , ansy = , ans = ;
struct point
{
int x , y ;
friend bool operator < (const point & a,const point & b)
{
return a.x < b.x || (a.x == b.x && a.y<b.y);
}
}; point p[maxn];
vector<int>Line; void initiation()
{
scanf("%d%d%d",&n,&w,&h);
Line.push_back();
Line.push_back(h);
for(int i = ; i < n ; ++ i)
{
scanf("%d%d",&p[i].x,&p[i].y);
Line.push_back(p[i].y);
}
} void solve()
{
sort(Line.begin() , Line.end());
int c = unique(Line.begin() , Line.end()) - Line.begin();
sort(p , p + n);
for(int i = ; i < c ; ++ i)
{
for(int j = i + ; j < c ; ++ j)
{
int L = Line[i];
int R = Line[j];
int h = R-L;
int pre = ;
for(int k = ; k < n ; ++ k)
{
if(p[k].y >= R || p[k].y <= L) continue;
int tx = p[k].x - pre;
int newans = min( tx , h );
if(newans > ans)
{
ans = newans;
ansx = pre;
ansy = L;
}
pre = p[k].x;
}
int tx = w - pre;
int newans = min( tx , h );
if(newans > ans)
{
ans = newans;
ansx = pre;
ansy = L;
}
}
}
printf("%d %d %d\n",ansx,ansy,ans);
} int main(int argc,char *argv[])
{
freopen("cricket.in","r",stdin);
freopen("cricket.out","w",stdout);
initiation();
if(n == ) printf("0 0 %d\n",min(w,h));
else solve();
return ;
}
Codeforces Gym 100002 C "Cricket Field" 暴力的更多相关文章
- Codeforces Gym 100002 E "Evacuation Plan" 费用流
"Evacuation Plan" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...
- Codeforces Gym 100002 Problem F "Folding" 区间DP
Problem F "Folding" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/ ...
- Codeforces Gym 100002 D"Decoding Task" 数学
Problem D"Decoding Task" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com ...
- Codeforces Gym 100002 B Bricks 枚举角度
Problem B Bricks" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 ...
- Codeforces Gym 100513M M. Variable Shadowing 暴力
M. Variable Shadowing Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513/ ...
- Codeforces Gym 100513G G. FacePalm Accounting 暴力
G. FacePalm Accounting Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513 ...
- Codeforces Gym 100342E Problem E. Minima 暴力
Problem E. MinimaTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/attac ...
- Codeforces Gym 100203G G - Good elements 暴力
G - Good elementsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...
- Codeforces Gym 101190M Mole Tunnels - 费用流
题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...
随机推荐
- liux vim命令
命令历史 以:和/开头的命令都有历史纪录,可以首先键入:或/然后按上下箭头来选择某个历史命令. 启动vim 在命令行窗口中输入以下命令即可 vim 直接启动vim vim filename 打开vim ...
- Solr DIH以Mysql为数据源批量创建索引
演示使用solr管理后台,以mysql为数据源,批量建索引的方法 测试于:Solr 4.5.1, mmseg4j 1.9.1, Jdk 1.6.0_45, Tomcat 6.0.37 | CentOS ...
- button捕捉回车键
为了给一个页面设置回车默认触发的按钮功能,我浏览了IE上的诸多方法,有的言不达意,有的读不懂,后来把高手的一段代码改造后,形成了一段代码,使这个问题的解决变得非常简章,有兴趣的朋友不妨一试.<s ...
- java web 学习十四(JSP原理)
一.什么是JSP? JSP全称是Java Server Pages,它和servle技术一样,都是SUN公司定义的一种用于开发动态web资源的技术. JSP这门技术的最大的特点在于,写jsp就像在写h ...
- Devexpress GridControl z
http://minmin86121.blog.163.com/blog/static/4968115720144194923578/ 1 AllowNullInput=False; --Devexp ...
- Delphi richedit获取选中文字
function TForm1.GetSendText(RichEdit: TExRichEdit): string;var MsgListInfo: TStrings; i, m, n: i ...
- 用javascript 面向对象制作坦克大战(四)
我们现在还差一个重要的功能,没错,敌人坦克的创建以及子弹击中敌人坦克时的碰撞检测功能. 5. 创建敌人坦克完成炮弹碰撞检测 5.1 创建敌人坦克对象 敌人坦克和玩家坦克一样,同样继承自我们的坦克 ...
- Vim小知识
在退出vim编辑的时候,强制退出是q! 感叹号在前,即!q,表示执行外部shell命令,感叹号在后,即q!,表示强制执行vi命令.
- SVM应用
我在项目中应用的SVM库是国立台湾大学林智仁教授开发的一套开源软件,主要有LIBSVM与LIBLINEAR两个,LIBSVM是对非线性数据进行分类,大家也比较熟悉,LIBLINEAR是对线性数据进行分 ...
- phonegap WebApp
打开网页浏览器,进入Android SDK网站(http://developer.android.com/sdk/index.html). 我们可以看到,Google官方提供了包括Windows平台在 ...