http://codeforces.com/contest/527/problem/C
C. Glass Carving
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular w mm × h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input
The first line contains three integers w, h, n (2 ≤ w, h ≤ 200 000, 1 ≤ n ≤ 200 000).
Next n lines contain the descriptions of the cuts. Each description has the form H y or V x. In the first case Leonid makes the horizontal cut at the distance y millimeters (1 ≤ y ≤ h - 1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance x (1 ≤ x ≤ w - 1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Examples
Input
Copy
4 3 4 H 2 V 2 V 3 V 1
Output
Copy
8 4 4 2
Input
Copy
7 6 5 H 4 V 3 V 5 H 2 V 1
Output
Copy
28 16 12 6 4
Note
Picture for the first sample test:
Picture for the second sample test:
线段树，未被切割过的行，列用0来表示，已经切割过的用1来表示，就是求行列的最长连续0个数
AC代码
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
#define maxn 200007
struct node1{
long long left=0;
long long right=0;
long long cnt=0;
long long flag=0;
};
struct node2{
long long left=0;
long long right=0;
long long cnt=0;
long long flag=0;
};
node1 nodes1[maxn<<2];
node2 nodes2[maxn<<2];
void pushup1(int rt)
{
if(nodes1[rt<<1].flag==0&&nodes1[rt<<1|1].flag==0)
{
nodes1[rt].left=nodes1[rt<<1].left+nodes1[rt<<1|1].left;
nodes1[rt].right=nodes1[rt].left;
nodes1[rt].cnt=0;
nodes1[rt].flag=0;
return;
}
if(nodes1[rt<<1].flag==0&&nodes1[rt<<1|1].flag==1)
{
nodes1[rt].left=nodes1[rt<<1].left+nodes1[rt<<1|1].left;
nodes1[rt].right=nodes1[rt<<1|1].right;
nodes1[rt].cnt=nodes1[rt<<1|1].cnt;
nodes1[rt].flag=1;
}
if(nodes1[rt<<1].flag==1&&nodes1[rt<<1|1].flag==0)
{
nodes1[rt].right=nodes1[rt<<1|1].right+nodes1[rt<<1].right;
nodes1[rt].left=nodes1[rt<<1].left;
nodes1[rt].cnt=nodes1[rt<<1].cnt;
nodes1[rt].flag=1;
}
if(nodes1[rt<<1].flag==1&&nodes1[rt<<1|1].flag==1)
{
nodes1[rt].left=nodes1[rt<<1].left;
nodes1[rt].right=nodes1[rt<<1|1].right;
nodes1[rt].cnt=max(nodes1[rt<<1].right+nodes1[rt<<1|1].left,max(nodes1[rt<<1].cnt,nodes1[rt<<1|1].cnt));
nodes1[rt].flag=1;
}
}
void pushup2(int rt)
{
if(nodes2[rt<<1].flag==0&&nodes2[rt<<1|1].flag==0)
{
nodes2[rt].left=nodes2[rt<<1].left+nodes2[rt<<1|1].left;
nodes2[rt].right=nodes2[rt].left;
nodes2[rt].cnt=0;
nodes2[rt].flag=0;
return;
}
if(nodes2[rt<<1].flag==0&&nodes2[rt<<1|1].flag==1)
{
nodes2[rt].left=nodes2[rt<<1].left+nodes2[rt<<1|1].left;
nodes2[rt].right=nodes2[rt<<1|1].right;
nodes2[rt].cnt=nodes2[rt<<1|1].cnt;
nodes2[rt].flag=1;
}
if(nodes2[rt<<1].flag==1&&nodes2[rt<<1|1].flag==0)
{
nodes2[rt].right=nodes2[rt<<1|1].right+nodes2[rt<<1].right;
nodes2[rt].left=nodes2[rt<<1].left;
nodes2[rt].cnt=nodes2[rt<<1].cnt;
nodes2[rt].flag=1;
}
if(nodes2[rt<<1].flag==1&&nodes2[rt<<1|1].flag==1)
{
nodes2[rt].left=nodes2[rt<<1].left;
nodes2[rt].right=nodes2[rt<<1|1].right;
nodes2[rt].cnt=max(nodes2[rt<<1].right+nodes2[rt<<1|1].left,max(nodes2[rt<<1].cnt,nodes2[rt<<1|1].cnt));
nodes2[rt].flag=1;
}
}
void build1(int l,int r,int rt)
{
if(l==r)
{
nodes1[rt].left=1;
nodes1[rt].right=1;
nodes1[rt].cnt=0;
nodes1[rt].flag=0;
return;
}
else
{
int mid=(l+r)>>1;
build1(l,mid,rt<<1);
build1(mid+1,r,rt<<1|1);
pushup1(rt);
}
}
void build2(int l,int r,int rt)
{
if(l==r)
{
nodes2[rt].left=1;
nodes2[rt].right=1;
nodes2[rt].cnt=0;
nodes2[rt].flag=0;
return;
}
else
{
int mid=(l+r)>>1;
build2(l,mid,rt<<1);
build2(mid+1,r,rt<<1|1);
pushup2(rt);
}
}
void update1(int L,int C,int l,int r,int rt)
{
if(l==r)
{
nodes1[rt].left=0;
nodes1[rt].right=0;
nodes1[rt].cnt=0;
nodes1[rt].flag=1;
return;
}
else
{
int mid=(l+r)>>1;
if(L<=mid)
update1(L,C,l,mid,rt<<1);
else
update1(L,C,mid+1,r,rt<<1|1);
pushup1(rt);
}
}
void update2(int L,int C,int l,int r,int rt)
{
if(l==r)
{
nodes2[rt].left=0;
nodes2[rt].right=0;
nodes2[rt].cnt=0;
nodes2[rt].flag=1;
return;
}
else
{
int mid=(l+r)>>1;
if(L<=mid)
update2(L,C,l,mid,rt<<1);
else
update2(L,C,mid+1,r,rt<<1|1);
pushup2(rt);
}
}
int main()
{
std::ios::sync_with_stdio(false);
long long w,h,n;
cin>>w>>h>>n;
build1(1,w,1);
build2(1,h,1);
for(int i=1;i<=n;++i)
{
char a;
int b;
cin>>a>>b;
if(a=='V')
{
update1(b,1,1,w,1);
long long temp1=max(nodes1[1].left+1,max(nodes1[1].cnt+1,nodes1[1].right));
long long temp2=max(nodes2[1].cnt+1,max(nodes2[1].left+1,nodes2[1].right));
cout<<min(temp1,w)*min(temp2,h)<<endl;
}
else
{
update2(b,1,1,h,1);
long long temp1=max(nodes1[1].cnt+1,max(nodes1[1].left+1,nodes1[1].right));
long long temp2=max(nodes2[1].cnt+1,max(nodes2[1].left+1,nodes2[1].right));
cout<<min(temp1,w)*min(temp2,h)<<endl;
}
}
return 0;
}