【ARC 063F】Snuke's Coloring 2

Description

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 (x_i,y_i).

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

• the region satisfying x<x_i within the rectangle
• the region satisfying x>x_i within the rectangle
• the region satisfying y<y_i within the rectangle
• the region satisfying y>y_i within the rectangle

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

题意：给定一个$W\times H$的二维平面，初始均为白色，有$n$个关键点$(x_{i},y_{i})$，对于每一个关键点选择一个方向，并将该方向上的所有网格涂成黑色。易得操作后白色部分一定是一个矩形，请最大化矩形周长。

分析：

观察可以得到一个性质，答案矩形一定会经过直线$x=\frac{W}{2}$或$y=\frac{H}{2}$。两种情况可以用相同的方式处理出答案。

将坐标离散化后，枚举矩形的上下边界，可以直接计算出矩形的左右边界。考虑用线段树进行优化。左右各开一个单调栈，在维护单调栈时在线段树上进行区间加减即可。（其实画图比较方便理解。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define LL long long
 #define lc(x) x<<1
 #define rc(x) x<<1|1
 using namespace std;
 const int N=3e5+;
 int w,h,n,ans,L,R;
 int mx[N*],tag[N*];
 struct node{int x,y;node(int _x=,int _y=):x(_x),y(_y){};}p[N],a[N],b[N];
 int read()
 {
     int x=,f=;char c=getchar();
     while(c<''||c>''){if(c=='-')f=-;c=getchar();}
     while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
     return x*f;
 }
 void modify(int x,int l,int r,int v)
 {
     if(L<=l&&r<=R){mx[x]+=v;tag[x]+=v;return;}
     int mid=(l+r)>>;
     if(L<=mid)modify(lc(x),l,mid,v);
     if(R>mid)modify(rc(x),mid+,r,v);
     mx[x]=max(mx[lc(x)],mx[rc(x)])+tag[x];
 }
 bool cmp(node a,node b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
 void work()
 {
     memset(mx,,sizeof(mx));
     memset(tag,,sizeof(tag));
     sort(p+,p+n+,cmp);
     int l=,r=;
     for(int i=;i<=n-;i++)
     {
         if(p[i].y<=h/)
         {
             int nxt=i-;
             while(l&&a[l].y<p[i].y)
             {
                 L=a[l].x;R=nxt;nxt=a[l].x-;
                 modify(,,n,a[l].y-p[i].y);l--;
             }
             if(nxt!=i-)a[++l]=node(nxt+,p[i].y);
         }
         else
         {
             int nxt=i-;
             while(r&&b[r].y>p[i].y)
             {
                 L=b[r].x;R=nxt;nxt=b[r].x-;
                 modify(,,n,p[i].y-b[r].y);r--;
             }
             if(nxt!=i-)b[++r]=node(nxt+,p[i].y);
         }
         a[++l]=node(i,);b[++r]=node(i,h);
         L=i;R=i;modify(,,n,h-p[i].x);
         ans=max(ans,mx[]+p[i+].x);
     }
 }
 int main()
 {
     w=read();h=read();n=read();
     for(int i=;i<=n;i++)p[i].x=read(),p[i].y=read();
     p[++n]=node(,);p[++n]=node(w,h);work();
     for(int i=;i<=n;i++)swap(p[i].x,p[i].y);
     swap(w,h);work();
     printf("%d",ans*);
     return ;
 }