BZOJ——1720: [Usaco2006 Jan]Corral the Cows 奶牛围栏

http://www.lydsy.com/JudgeOnline/problem.php?id=1720

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 177  Solved: 90
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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*
+------

HINT

 

Source

Gold

求最小的代价，考虑二分（logmaxlen）
发现数据范围支持n^2logmaxlen的复杂度
现将所求正方形看做是一个无限高的矩形，
O(n)枚举一个右端点，确定出左端点后，
再O(n)判断在规定的高度内能否得到C个糖果

 #include <algorithm>
 #include <cstdio>

 inline void read(int &x)
 {
     x=; register char ch=getchar();
     for(; ch>''||ch<''; ) ch=getchar();
     for(; ch>=''&&ch<=''; ch=getchar()) x=x*+ch-'';
 }
 const int N();
 int c,n;
 struct Node {
     int x,y;
     bool operator < (const Node&a)const
     {
         return x<a.x;
     }
 }pos[N];

 int L,R,Mid,ans,cnt,tmp[N];
 inline bool judge(int l,int r)
 {
     if(r-l+<c) return ; cnt=;
     for(int i=l; i<=r; ++i) tmp[++cnt]=pos[i].y;
     std::sort(tmp+,tmp+cnt+);
     for(int i=c; i<=cnt; ++i)
         if(tmp[i]-tmp[i-c+]<=Mid) return ;
     return ;
 }
 inline bool check(int x)
 {
     int l=,r=;
     for(; r<=n; ++r)
     {
         if(pos[r].x-pos[l].x>x)
             if(judge(l,r-)) return ;
         for(; pos[r].x-pos[l].x>x; ) l++;
     }
     return judge(l,n);
 }

 int Presist()
 {
     read(c),read(n);
     for(int i=; i<=n; ++i)
         read(pos[i].x),read(pos[i].y);
     std::sort(pos+,pos++n);
     for(R=1e4; L<=R; )
     {
         Mid=L+R>>;
         if(check(Mid))
         {
             R=Mid-;
             ans=Mid+;
         }
         else L=Mid+;
     }
     printf("%d\n",ans);
     return ;
 }

 int Aptal=Presist();
 int main(int argc,char**argv){;}