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

 1 #include <algorithm>
 2 #include <cstdio>
 3 
 4 inline void read(int &x)
 5 {
 6     x=0; register char ch=getchar();
 7     for(; ch>'9'||ch<'0'; ) ch=getchar();
 8     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
 9 }
10 const int N(523);
11 int c,n;
12 struct Node {
13     int x,y;
14     bool operator < (const Node&a)const
15     {
16         return x<a.x;
17     }
18 }pos[N];
19 
20 int L,R,Mid,ans,cnt,tmp[N];
21 inline bool judge(int l,int r)
22 {
23     if(r-l+1<c) return 0; cnt=0;
24     for(int i=l; i<=r; ++i) tmp[++cnt]=pos[i].y;
25     std::sort(tmp+1,tmp+cnt+1);
26     for(int i=c; i<=cnt; ++i)
27         if(tmp[i]-tmp[i-c+1]<=Mid) return 1;
28     return 0;
29 }
30 inline bool check(int x)
31 {
32     int l=1,r=1;
33     for(; r<=n; ++r)
34     {
35         if(pos[r].x-pos[l].x>x)
36             if(judge(l,r-1)) return 1;
37         for(; pos[r].x-pos[l].x>x; ) l++;
38     }
39     return judge(l,n);
40 }
41 
42 int Presist()
43 {
44     read(c),read(n);
45     for(int i=1; i<=n; ++i)
46         read(pos[i].x),read(pos[i].y);
47     std::sort(pos+1,pos+1+n);
48     for(R=1e4; L<=R; )
49     {
50         Mid=L+R>>1;
51         if(check(Mid))
52         {
53             R=Mid-1;
54             ans=Mid+1;
55         }
56         else L=Mid+1;
57     }
58     printf("%d
",ans);
59     return 0;
60 }
61 
62 int Aptal=Presist();
63 int main(int argc,char**argv){;}