【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 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.

    约翰打算建一个围栏来圈养他的奶牛．作为最挑剔的兽类，奶牛们要求这个围栏必须是正方形的，而且围栏里至少要有C(1≤C≤500)个草场，来供应她们的午餐．

    约翰的土地上共有N(C≤N≤500)个草场，每个草场在一块lxl的方格内，而且这个方格的坐标不会超过10000.有时候，会有多个草场在同一个方格内，那他们的坐标就会相同．

    告诉约翰，最小的围栏的边长是多少？

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.

    第1行输入C和N，接下来N行每行输入一对整数，表示一个草场所在方格的坐标

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.

    输入最小边长．

Sample Input

3 4
1 2
2 1
4 1
5 2

Sample Output

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

|CCCC
|CCCC
|*CCC*
|C*C*
+------

题解：二维双指针法的完美结合

双指针法就是令l=1，从1到n枚举右指针r，然后始终保证[l,r]的区间是满足题目要求的区间，不满足就使l++，并每次用[l,r]更新答案（感觉就是简化的单调队列）

如果坐标都是一维的，我们只用双指针法就能搞定，但是由于是二维的，所以我们要用两重双指针法（四指针法。。。）

先二分答案，分别用双指针法维护纵坐标和横坐标，具体还是看代码吧~

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,m,nx,ny;
struct field
{
	int x,y;
}p[510];
int rx[510],ry[510],s[510];
bool cmp1(field a,field b)
{
	return a.x<b.x;
}
bool cmp2(field a,field b)
{
	return a.y<b.y;
}
bool solve(int ml)
{
	int i,a,b,c,d,sc,sd;
	a=b=0;
	memset(s,0,sizeof(s));
	while(b<n&&rx[p[b+1].x]-rx[1]+1<=ml)	s[p[++b].y]++;
	for(;b<=n;s[p[++b].y]++)
	{
		while(rx[p[b].x]-rx[p[a+1].x]+1>ml)	s[p[++a].y]--;
		c=d=sc=sd=0;
		while(d<ny&&ry[d+1]-ry[1]+1<=ml)	sd+=s[++d];
		for(;d<=ny;sd+=s[++d])
		{
			while(ry[d]-ry[c+1]+1>ml)	sc+=s[++c];
			if(sd-sc>=m)	return true;
		}
	}
	return false;
}
int main()
{
	scanf("%d%d",&m,&n);
	int i;
	rx[0]=ry[0]=-1;
	for(i=1;i<=n;i++)	scanf("%d%d",&p[i].x,&p[i].y);
	sort(p+1,p+n+1,cmp2);
	for(i=1;i<=n;i++)
	{
		if(p[i].y>ry[ny])	ry[++ny]=p[i].y;
		p[i].y=ny;
	}
	sort(p+1,p+n+1,cmp1);
	for(i=1;i<=n;i++)
	{
		if(p[i].x>rx[nx])	rx[++nx]=p[i].x;
		p[i].x=nx;
	}
	int l=1,r=max(rx[nx],ry[ny]),mid;
	while(l<r)
	{
		mid=l+r>>1;
		if(solve(mid))	r=mid;
		else	l=mid+1;
	}
	printf("%d",r);
	return 0;
}