Codeforces Round #296 (Div. 2) C. Glass Carving

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 wmm  ×  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

4 3 4
H 2
V 2
V 3
V 1

output

8
4
4
2

input

7 6 5
H 4
V 3
V 5
H 2
V 1

output

28
16
12
6
4

Note

Picture for the first sample test: 

Picture for the second sample test: 

借鉴自：http://blog.csdn.net/acvay/article/details/44428235

题意简单，题目半天写不出来。找了题解原来是stl的运用。学习一下吧，没想到set效率这么高。

首先用set存在水平和竖直方向所存在的边。随着切割的过程，逐渐把边加到相应集合里。在加边的过程中，用wi和hi存储水平和竖直方向上两边之间夹的最大宽度，加边时，相应修改两边之间最大宽度即可。最后结果当然是水平和竖直方向最大宽度乘积。

PS.其实做法十分暴力，主要学习用法吧，感觉自己越来越弱……

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define INF 0x3f3f3f3
const int N=200005;
const int mod=1e9+7;

set<int>::iterator i,j;
set<int> ve,ho;
int wi[N],hi[N];

void cut(set<int> &s, int *a,int p)
{
    s.insert(p);
    i=j=s.find(p);
    --i,++j;--a[*j-*i];
    ++a[p-*i],++a[*j-p];
}

int main(){
    int mw,mh,p,w,h,n;
    char temp;
    while (cin>>w>>h>>n) {
        memset(wi, 0, sizeof(wi));
        memset(hi, 0, sizeof(hi));
        ve.clear();
        ho.clear();
        ve.insert(0),ve.insert(w);
        ho.insert(0),ho.insert(h);
        getchar();
        mw=w;
        mh=h;
        wi[w]=1;
        hi[h]=1;
        for (int i=0; i<n; i++) {
            scanf("%c %d",&temp,&p);
            getchar();
            if (temp=='V') cut(ve,wi,p);
            else cut(ho,hi,p);
            while (!wi[mw]) mw--;
            while (!hi[mh]) mh--;
            printf("%I64d\n",(long long)mw*mh);
        }
    }
    return 0;
}