妙笔难书一纸愁肠，苍白的誓言，终究抵不过岁月的遗忘。

题目描述

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.

输入

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.

输出

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.

样例输入

复制样例数据

样例输出

提示

|* *
| * *
+------
Below is one 4x4 solution (C's show most of the corral's area); many others exist.
|CCCC
|CCCC
|*CCC*
|C*C*

+------

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
int C,N,mid,jg,tp[10010];
struct node
{int x;int y;
} p[10010];
int cmp(node a,node b)
{return a.x<b.x;
}
int qz(int l,int r)
{if(r-l+1<C)return 0;int ct=0;for(int i=l; i<=r; i++)tp[++ct]=p[i].y;sort(tp+1,tp+ct+1);for(int i=C; i<=ct; i++)if(tp[i]-tp[i-C+1]<=mid)return 1;return 0;
}
int pd(int x)
{int l=1,i;for(int i=1; i<=N; i++){if(p[i].x-p[l].x>x){if(qz(l,i-1))return 1;while(p[i].x-p[l].x>x)l++;}}if(qz(l,N))return 1;return 0;
}
int main()
{
//    cin>>C>>N;scanf("%d%d",&C,&N);for(int i=1; i<=N; i++)scanf("%d%d",&p[i].x,&p[i].y);sort(p+1,p+N+1,cmp);int l=0,r=10000;while(l<=r){mid=(l+r)/2;if(pd(mid)){jg=mid+1;r=mid-1;}elsel=mid+1;}printf("%d\n",jg);return 0;
}

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妙笔难书一纸愁肠，苍白的誓言，终究抵不过岁月的遗忘。

样例输入

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