C. Glass Carving
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
inputCopy
4 3 4
H 2
V 2
V 3
V 1
outputCopy
8
4
4
2
inputCopy
7 6 5
H 4
V 3
V 5
H 2
V 1
outputCopy
28
16
12
6
4
Note
Picture for the first sample test:

Picture for the second sample test:

set的用法，很明显我们每次删除一个东西，然后插入一个东西，我们用平衡树来解决。

然后最大的面积，就是 最大的横向长度*最大的纵向长度 。

用set维护切割的位置，我们这样就可以找到当前切割的长度，以及左右两边切割的位置，然后multiset维护数值即可。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+10;
int w,h,n;
set<int> sw,sh;
multiset<int> mw,mh;
inline void cut(set<int> &s,multiset<int> &ms,int x){
	s.insert(x);	auto l=s.find(x);	auto r=l;	l--,r++;
	ms.erase(ms.find(*r-*l)); ms.insert(*r-x); ms.insert(x-*l);
}
signed main(){
	cin>>w>>h>>n;
	sw.insert(0); sw.insert(w); sh.insert(0); sh.insert(h);
	mw.insert(w); mh.insert(h);
	while(n--){
		char op[2];	int x;	scanf("%s %lld",op,&x);
		if(op[0]=='H')	cut(sh,mh,x);
		else	cut(sw,mw,x);
		printf("%lld\n",(*--mw.end())*(*--mh.end()));
	}
	return 0;
}