Corral the Cows
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 1352   Accepted: 565

Description

Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain at least C (1 <= C <= 500) clover fields for afternoon treats. The corral's edges must be parallel to the X,Y axes.

FJ's land contains a total of N (C <= N <= 500) clover fields, each a block of size 1 x 1 and located at with its lower left corner at integer X and Y coordinates each in the range 1..10,000. Sometimes more than one clover field grows at the same location; such a field would have its location appear twice (or more) in the input. A corral surrounds a clover field if the field is entirely located inside the corral's borders.

Help FJ by telling him the side length of the smallest square containing C clover fields.

Input

Line 1: Two space-separated integers: C and N

Lines 2..N+1: Each line contains two space-separated integers that are the X,Y coordinates of a clover field.

Output

Line 1: A single line with a single integer that is length of one edge of the minimum size square that contains at least C clover fields.

Sample Input

3 4
1 2
2 1
4 1
5 2

Sample Output

4

Hint

Explanation of the sample:

|*   *

| * *

+------

Below is one 4x4 solution (C's show most of the corral's area); many others exist.

|CCCC

|CCCC

|*CCC*

|C*C*

+------

Source

 
 
题意:需要你搭建一个正方形的围栏,围栏里面需要有至少C个特殊的点,问你正方形最短的边长是多少
题解:给的坐标可能会重复,所以我们按照x坐标的和y坐标从小到大排序后离散化,用二维前缀和记录区域内点的个数,然后二分答案,每次检查时检查区域内的点是否满足大于C个即可
代码如下:
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = ;
int C,n;
struct node{
int x,y;
}p[maxn];
int xn,yn,rx[maxn],ry[maxn];
int sum[maxn][maxn];
bool cmp1(node a,node b){
return a.x<b.x;
}
bool cmp2(node a,node b){
return a.y<b.y;
}
bool check(int k){
for(int a=,b=;;a++){
while(rx[b+]-rx[a]+<=k&&b<xn) b++;
for(int c=,d=;;c++){
while(ry[d+]-ry[c]+<=k&&d<yn) d++;
int ans=sum[b][d]+sum[a-][c-]-sum[a-][d]-sum[b][c-];
if(ans>=C) return true;
if(d==yn) break;
}
if(b==xn) break;
}
return false;
} int main(){
scanf("%d%d",&C,&n);
for(int i=;i<=n;i++){
scanf("%d%d",&p[i].x,&p[i].y);
}
sort(p+,p+n+,cmp1);
xn=;rx[]=p[].x;p[].x=;
for(int i=;i<=n;i++){
if(p[i].x!=p[i-].x) rx[++xn]=p[i].x;
p[i].x=xn;
}
sort(p+,p+n+,cmp2);
yn=;ry[]=p[].y;p[].y=;
for(int i=;i<=n;i++){
if(p[i].y!=p[i-].y) ry[++yn]=p[i].y;
p[i].y=yn;
}
for(int i=;i<=n;i++) sum[p[i].x][p[i].y]++;
for(int i=;i<=xn;i++){
for(int j=;j<=yn;j++){
sum[i][j]+=sum[i][j-]+sum[i-][j]-sum[i-][j-];
}
}
int l=,r=max(rx[xn],ry[yn]),ans;
while(l<=r){
int mid=l+r>>;
if(check(mid)){
ans=mid;
r=mid-;
}else{
l=mid+;
}
}
printf("%d\n",ans);
}

POJ 3179 Corral the Cows的更多相关文章

  1. 洛谷 P2862 [USACO06JAN]把牛Corral the Cows 解题报告

    P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being fi ...

  2. POJ 2387 Til the Cows Come Home (图论,最短路径)

    POJ 2387 Til the Cows Come Home (图论,最短路径) Description Bessie is out in the field and wants to get ba ...

  3. 【BZOJ1720】[Usaco2006 Jan]Corral the Cows 奶牛围栏 双指针法

    [BZOJ1720][Usaco2006 Jan]Corral the Cows 奶牛围栏 Description Farmer John wishes to build a corral for h ...

  4. POJ.2387 Til the Cows Come Home (SPFA)

    POJ.2387 Til the Cows Come Home (SPFA) 题意分析 首先给出T和N,T代表边的数量,N代表图中点的数量 图中边是双向边,并不清楚是否有重边,我按有重边写的. 直接跑 ...

  5. 洛谷——P2862 [USACO06JAN]把牛Corral the Cows

    P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being fi ...

  6. 洛谷P2862 [USACO06JAN]把牛Corral the Cows

    P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being fi ...

  7. 洛谷 P2862 [USACO06JAN]把牛Corral the Cows

    P2862 [USACO06JAN]把牛Corral the Cows 题目描述 Farmer John wishes to build a corral for his cows. Being fi ...

  8. 【POJ 3179】 Corral the Cows

    [题目链接] http://poj.org/problem?id=3179 [算法] 首先,我们发现答案是具有单调性的,也就是说,如果边长为C的正方形可以,那么比边长C大的正方形也可以,因此,可以二分 ...

  9. POJ 2387 Til the Cows Come Home

    题目链接:http://poj.org/problem?id=2387 Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K ...

随机推荐

  1. ABAP CDS ON HANA-(11)ABAP CDSでの関連付け

    Association in ABAP CDS An association in CDS view joins different data sources. Defining and using ...

  2. [Python 3.X]python练习笔记[2]-----用python实现七段数码管显示年月日

    #SevenDigitsDrawV2.py import turtle import time def drawGap(i):#绘制数码管间隔 turtle.penup() turtle.fd(i) ...

  3. 【python3.X】python练习笔记[1]

    ##三位数水仙花 ##方法一,小于指定数字的水仙花数 x=eval(input()) a,b,c = 0,0,0 for i in range (100,x,1): a=i%10 b=i//100 c ...

  4. CentOS7 添加自定义系统服务案例

    示例一: 执行脚本/root/project/systemctl/test.sh() ######################################################### ...

  5. 了解url

    我对自己知道关于url的编码和解码的一些进行了一下整理. 我们的例子是百度翻译的地址: http://fanyi.baidu.com/translate#en/zh/The%20%22%3F%20ar ...

  6. Linux-Ps命令使用

    ps -ef | grep java ps aux | grep java ps aux 是用BSD的格式来显示Java进程 显示的项目有: USER        PID %CPU %MEM    ...

  7. python接口测试(一)——http请求及token获取

    使用python对当前的接口进行简单的测试 1.接口测试是针对软件对外提供服务得接口得输入输出进行得测试,验证接口功能与接口描述文档得一致性 返回结果可以为字符串,json,xml等 2.接口的请求方 ...

  8. python学习总结----异常处理

    相关概念 - 错误:程序运行之前的语法错误,如:关键字.缩进不齐.括号不成对. - 异常:在程序运行过程中出现的问题,如:除数为0.对象属性不存在等. 异常处理 - 说明:异常处理可以理解为特殊的流程 ...

  9. linux常用命令补充详细

    1.ls命令 就是list的缩写,通过ls 命令不仅可以查看linux文件夹包含的文件,而且可以查看文件权限(包括目录.文件夹.文件权限)查看目录信息等等 常用参数搭配: ls -a 列出目录所有文 ...

  10. Mysql性能优化四:分库,分区,分表,你们如何做?

    分库分区分表概念 分区 就是把一张表的数据分成N个区块,在逻辑上看最终只是一张表,但底层是由N个物理区块组成的 分表 就是把一张数据量很大的表按一定的规则分解成N个具有独立存储空间的实体表.系统读写时 ...