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≦iN) point was (xi,yi).

Then, for each 1≦iN, 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≦xiW (1≦iN)
  • 0≦yiH (1≦iN)
  • WH (21:32, added), xi and yi are integers.
  • If ij, then xixj and yiyj.

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
10 10 4
1 6
4 1
6 9
9 4

Sample Output 1

Copy
32

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

AtCoder Regular Contest 063 F : Snuke’s Coloring 2 利用性质 单调栈 线段树_第1张图片

Sample Input 2

Copy
5 4 5
0 0
1 1
2 2
4 3
5 4

Sample Output 2

Copy
12

Sample Input 3

Copy
100 100 8
19 33
8 10
52 18
94 2
81 36
88 95
67 83
20 71

Sample Output 3

Copy
270

Sample Input 4

Copy
100000000 100000000 1
3 4

Sample Output 4

Copy
399999994

AtCoder Regular Contest 063 F : Snuke’s Coloring 2 利用性质 单调栈 线段树_第2张图片


#include
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#include
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=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<

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