算法讨论：

先按照y坐标排好序。二分答案，利用线段树判断是否可行。

这算不算是离散化呢？算是广义的离散化吧。

开始的时候没有更新，后来意识到错误后加上了更新反而错了，于是就偷了个懒把线段树build了log10000次，这样时间复杂度就变成了（log10000）^2*10000,过得很纠结……

其实不用build那么多次线段树的。

代码：

program corral;//By_Thispoetconst	maxn=505;maxx=10005;var	i,j,p,c,n,mid,left,right,ans,k	:longint;	tree							:array[0..maxx*10]of record l,r,num:longint;end;	x,y								:array[0..maxn]of longint;procedure swap(var i,j:longint);begin	if i<>j then begin i:=i xor j;j:=i xor j;i:=i xor j;end;end;procedure qsort(l,r:longint);var i,j,k:longint;begin	i:=l;j:=r;k:=y[i+random(j-i+1)];	repeat		while y[i]
     
      k do dec(j);		if i<=j then begin			swap(x[i],x[j]);swap(y[i],y[j]);inc(i);dec(j);		end;	until i>j;	if l
      
       =r then exit;mid:=(l+r)>>1;	build_tree(code*2,l,mid);build_tree(code*2+1,mid,r);end;procedure insert_tree(code,l,r,data:longint);var mid:longint;begin	if (tree[code].l=l)and(tree[code].r=r) then begin		inc(tree[code].num,data);exit;	end;mid:=(tree[code].l+tree[code].r)>>1;	if mid>=r then insert_tree(code*2,l,r,data) else insert_tree(code*2+1,l,r,data);	tree[code].num:=tree[code*2].num+tree[code*2+1].num;end;function count(code,l,r:longint):longint;var mid:longint;begin	if (tree[code].l=l)and(tree[code].r=r) then exit(tree[code].num);	mid:=(tree[code].l+tree[code].r)>>1;	if mid>=r then exit(count(code*2,l,r)) else if mid<=l then exit(count(code*2+1,l,r));	exit(count(code*2,l,mid)+count(code*2+1,mid,r));end;function min(i,j:longint):longint;begin	if i
       
        =c then begin			for j:=i to p do if count(1,x[j]-1,min(x[j]+pos-1,maxx))>=c then exit(true);		end;		insert_tree(1,x[i]-1,x[i],-1);	end;	exit(false);end;begin	readln(c,n);randomize;	for i:=1 to n do readln(x[i],y[i]);qsort(1,n);p:=0;	left:=0;right:=maxx;	while left<=right do begin		mid:=(left+right)>>1;		build_tree(1,0,maxx);		if check(mid) then begin			right:=mid-1;ans:=mid;		end else left:=mid+1;	end;	writeln(ans);end.