Codeforces Round #296 (Div. 2) C(STL_set)

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.

Sample test(s)

input

4 3 4H 2V 2V 3V 1

output

8442

input

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

output

28161264

Note

Picture for the first sample test:

Picture for the second sample test:

题意：在一个w*h的矩形里，进行n次切割，其中H是横切，W是竖切，求每次切完后最大的矩形面积。

思路：屌屌哒set应用，真是给跪了。记得用long long

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;set<LL >W,H;//记录边的位置set<LL >::iterator i,j;//记录当前插入边的前后位置LL WW[200010],HH[200010];//记录存在的边的边长值void cut(set<LL >&s,LL a[],LL key){    s.insert(key);    i=j=s.find(key);    i--;    j++;    --a[*j-*i];//除掉被分开的长或宽    ++a[key-*i];//产生新的长或宽    ++a[*j-key];//同上}int main(){    LL w,h,n;    char str[5];    LL key;    while(~scanf("%lld %lld %lld",&w,&h,&n)) {        memset(WW,0,sizeof(WW));        memset(HH,0,sizeof(HH));        W.clear();        H.clear();        W.insert(0);        W.insert(w);        H.insert(0);        H.insert(h);        WW[w]=HH[h]=1;        WW[0]=HH[0]=1;        LL wi=w;        LL hi=h;        while(n--) {            scanf("%s %lld",str,&key);            if(str[0]=='V')                cut(W,WW,key);            else                cut(H,HH,key);            while(!WW[wi])                --wi;            while(!HH[hi])                --hi;            printf("%lld\n",wi*hi);        }    }    return 0;}