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
10 10 4 1 6 4 1 6 9 9 4
Sample Output 1
32
In this case, the maximum perimeter of 32 can be obtained by painting the rectangle as follows:
Sample Input 2
5 4 5 0 0 1 1 2 2 4 3 5 4
Sample Output 2
12
Sample Input 3
100 100 8 19 33 8 10 52 18 94 2 81 36 88 95 67 83 20 71
Sample Output 3
270
Sample Input 4
100000000 100000000 1 3 4
Sample Output 4
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; }