【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<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.
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。
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 }
