AtCoder Regular Contest 063 F : Snuke’s Coloring 2 利用性质单调栈线段树

F - すぬけ君の塗り絵 2 / Snuke's Coloring 2

Time limit : 4sec / Memory limit : 256MB

Score : 1600 points

Problem Statement

There is a rectangle in the xy-plane, with its lower left corner at (0,0) and its upper right corner at (W,H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white.

Snuke plotted N points into the rectangle. The coordinate of the i-th (1≦i≦N) point was (xi,yi).

Then, for each 1≦i≦N, he will paint one of the following four regions black:

the region satisfying x<xi within the rectangle
the region satisfying x>xi within the rectangle
the region satisfying y<yi within the rectangle
the region satisfying y>yi within the rectangle

Find the longest possible perimeter of the white region of a rectangular shape within the rectangle after he finishes painting.

Constraints

1≦W,H≦108
1≦N≦3×105
0≦xi≦W (1≦i≦N)
0≦yi≦H (1≦i≦N)
W, H (21:32, added), xi and yi are integers.
If i≠j, then xi≠xj and yi≠yj.

Input

The input is given from Standard Input in the following format:

W H N
x1 y1
x2 y2
:
xN yN

Output

Print the longest possible perimeter of the white region of a rectangular shape within the rectangle after Snuke finishes painting.

Sample Input 1

Copy

Sample Output 1

Copy

In this case, the maximum perimeter of 32 can be obtained by painting the rectangle as follows:

Sample Input 2

Copy

Sample Output 2

Copy

Sample Input 3

Copy

Sample Output 3

Copy

Sample Input 4

Copy

100000000 100000000 1
3 4

Sample Output 4

Copy

399999994

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<map>
#include<set>
using namespace std;

typedef long long ll;

inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=10*x+ch-'0';ch=getchar();}
	return x*f;
}
void print(int x)
{if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}

const int N=300100;

int w,h,n;

struct point
{
	int x,y;
	
	friend bool operator<(const point &a,const point &b)
	{return a.x<b.x;}
	
}p[N];

struct segment_tree{int w,tag;}tr[N<<2];

inline void pushup(int k)
{tr[k].w=max(tr[k<<1].w,tr[k<<1|1].w);}

inline void pushdown(int k)
{
	if(tr[k].tag)
	{
		int tmp(tr[k].tag);
		tr[k<<1].w+=tmp,tr[k<<1].tag+=tmp,
		tr[k<<1|1].w+=tmp,tr[k<<1|1].tag+=tmp,
		tr[k].tag=0;
	}
}

inline void modify(int k,int l,int r,int x,int y,int val)
{
	if(l>=x && r<=y)
	{tr[k].w+=val,tr[k].tag+=val;return ;}
	int mid=(l+r)>>1;pushdown(k);
	if(x<=mid) modify(k<<1,l,mid,x,y,val);
	if(y>mid) modify(k<<1|1,mid+1,r,x,y,val);
	pushup(k);
}

int query(int k,int l,int r,int x)
{
	if(l==r) return tr[k].w;
	int mid=(l+r)>>1;pushdown(k);
	return x<=mid ? query(k<<1,l,mid,x) : query(k<<1|1,mid+1,r,x);
}

int Y[N];

int st[2][N],top[2];

int ans(0);

void solve()
{
	register int i;
	sort(p+2,p+1+n);
	
	top[0]=top[1]=0,
	st[0][0]=st[1][0]=1;
	modify(1,1,n,1,n,h<<1);
	for(i=2;i<=n;++i)
	{
		modify(1,1,n,1,i-1,(p[i].x-p[i-1].x)<<1);
		ans=max(ans,tr[1].w);
		if(p[i].y>=(h>>1))
		{
			modify(1,1,n,st[0][top[0]],i-1,(p[i].y-h)<<1);
			while(top[0] && p[st[0][top[0]]].y>=p[i].y)
				modify(1,1,n,st[0][top[0]-1],st[0][top[0]]-1,(p[i].y-p[st[0][top[0]]].y)<<1),
				top[0]--;
			st[0][++top[0]]=i;
		}
		else
		{
			modify(1,1,n,st[1][top[1]],i-1,-p[i].y<<1);
			while(top[1] && p[st[1][top[1]]].y<=p[i].y)
				modify(1,1,n,st[1][top[1]-1],st[1][top[1]]-1,(p[st[1][top[1]]].y-p[i].y)<<1),
				top[1]--;
			st[1][++top[1]]=i;
		}
	}
	memset(tr,0,sizeof(tr));
}

int main()
{
	w=read(),h=read(),n=read()+1;
	register int i;
	for(i=2;i<=n;++i)
		p[i].x=read(),p[i].y=read();
	p[++n]=(point){w,h};
	
	solve();
	swap(w,h);
	for(i=1;i<=n;++i)
		swap(p[i].x,p[i].y);
	solve();
	cout<<ans<<endl;
	return 0;
}