【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 (xi,yi).

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

  • the region satisfying x<xwithin the rectangle
  • the region satisfying x>xi within the rectangle
  • the region satisfying y<yi within the rectangle
  • the region satisfying y>ywithin 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}$$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。

 

 1 #include<cstdio>
 2 #include<algorithm> 
 3 #include<cstring>
 4 #define LL long long
 5 #define lc(x) x<<1
 6 #define rc(x) x<<1|1
 7 using namespace std;
 8 const int N=3e5+5;
 9 int w,h,n,ans,L,R;
10 int mx[N*4],tag[N*4];
11 struct node{int x,y;node(int _x=0,int _y=0):x(_x),y(_y){};}p[N],a[N],b[N]; 
12 int read()
13 {
14     int x=0,f=1;char c=getchar();
15     while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
16     while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
17     return x*f;
18 }
19 void modify(int x,int l,int r,int v)
20 {
21     if(L<=l&&r<=R){mx[x]+=v;tag[x]+=v;return;}
22     int mid=(l+r)>>1;
23     if(L<=mid)modify(lc(x),l,mid,v);
24     if(R>mid)modify(rc(x),mid+1,r,v);
25     mx[x]=max(mx[lc(x)],mx[rc(x)])+tag[x];
26 }
27 bool cmp(node a,node b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
28 void work()
29 {
30     memset(mx,0,sizeof(mx));
31     memset(tag,0,sizeof(tag));
32     sort(p+1,p+n+1,cmp);
33     int l=0,r=0;
34     for(int i=1;i<=n-1;i++)
35     {
36         if(p[i].y<=h/2)
37         {
38             int nxt=i-1;
39             while(l&&a[l].y<p[i].y)
40             {
41                 L=a[l].x;R=nxt;nxt=a[l].x-1;
42                 modify(1,1,n,a[l].y-p[i].y);l--; 
43             }
44             if(nxt!=i-1)a[++l]=node(nxt+1,p[i].y);
45         }
46         else
47         {
48             int nxt=i-1;
49             while(r&&b[r].y>p[i].y)
50             {
51                 L=b[r].x;R=nxt;nxt=b[r].x-1;
52                 modify(1,1,n,p[i].y-b[r].y);r--;
53             }
54             if(nxt!=i-1)b[++r]=node(nxt+1,p[i].y);
55         }
56         a[++l]=node(i,0);b[++r]=node(i,h);
57         L=i;R=i;modify(1,1,n,h-p[i].x);
58         ans=max(ans,mx[1]+p[i+1].x);
59     }
60 }
61 int main()
62 {
63     w=read();h=read();n=read();
64     for(int i=1;i<=n;i++)p[i].x=read(),p[i].y=read();
65     p[++n]=node(0,0);p[++n]=node(w,h);work();
66     for(int i=1;i<=n;i++)swap(p[i].x,p[i].y);
67     swap(w,h);work();
68     printf("%d",ans*2);
69     return 0;
70 }
View Code

 

posted @ 2018-04-22 14:45  Zsnuo  阅读(537)  评论(0编辑  收藏  举报