其实可以不转45度直接设一次方程解

这里有 sub2 sub3 sub4的全部做法。

#include<bits/stdc++.h>

using namespace std;

#define int long long

int aBs(int x){
	if (x<0) return -x;
	return x;
}

int n,m;
struct node{
	int x,y;
}a[1000006];

void main_solve1(){ //sub2
	int anss=m+1;
	for (int s=1;s<=m;s++){
		int l=1,r=m,ans=m+1;
		while (l<=r){
			int mid=(l+r)>>1;
			
			int L=1,R=m;
			for (int i=1;i<=n;i++){
				if (aBs(a[i].x-a[i].y)<=mid) continue;
				int diss=mid-aBs(s-a[i].x);
				L=max(L,a[i].y-diss);
				R=min(R,a[i].y+diss);
			}
			if (L<=R) r=mid-1,ans=mid;
			else l=mid+1;
		}
		anss=min(anss,ans);
	}
	printf("%d\n",anss);
	
	return;
}
void main_solve2(){ //sub3
	int ans=(aBs(a[1].x-a[2].x)+aBs(a[1].y-a[2].y)-1)/2 +1;
	ans=min(ans, aBs(a[1].x-a[1].y));
	ans=min(ans, aBs(a[2].x-a[2].y));
	printf("%lld\n",ans);
	return;
}
void main_solve(){ //正解
	int l=1,r=m,ans=m+1;
	while (l<=r){
		int mid=(l+r)>>1;
		
		int xmin=0xc0c0c0c0c0c0c0c0,xmax=0x3f3f3f3f3f3f3f3f;
		int ymin=0xc0c0c0c0c0c0c0c0,ymax=0x3f3f3f3f3f3f3f3f;
		for (int i=1;i<=n;i++){
			if (a[i].y-a[i].x<=mid) continue;
			ymax=min(ymax, a[i].x-a[i].y+mid);
			ymin=max(ymin, a[i].x-a[i].y-mid);
			
			xmax=min(xmax, a[i].x+a[i].y+mid);
			xmin=max(xmin, a[i].x+a[i].y-mid);
		}
		if (xmin<=xmax && ymin<=ymax){
			r=mid-1;
			ans=mid;
		}
		else{
			l=mid+1; 
		}
	}
	printf("%lld\n",ans);
	return;
}

signed main(){
//	freopen("A.in","r",stdin);
	scanf("%lld%lld",&n,&m);
	for (int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i].x,&a[i].y);
		if (a[i].x>a[i].y) swap(a[i].x,a[i].y);
	}
	if (n==2){
		main_solve2();
		return 0;
	}
	if (n<=1000 && m<=1000){
		main_solve1();
		return 0;
	}
	main_solve();
	return 0;
}