CodeForces 527C 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.

Sample test(s)

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:

题意：切玻璃，切一刀横的或者竖的，问当前玻璃块中最大的面积是多大

用STL里的Set，内部是用红黑树实现的，复杂度logn。

Set记录当前切割点位置。

开两个数组记录横的和竖的线段数量。

最后每次倒着从最大值枚举长度，判断数组里有没有这么长的线段，那么最先找到的就是最大的长和宽，乘起来就行了。

需要注意的地方。最后的面积结果可能会超int，所以用long long.

数组记录线段的时候，不能直接把没有的置为0，有的置为1，因为线段中可能相同长度的有多条。应该用自增自减的方式记录长度++,--.

#include 
#include 
#include 

using namespace std;
const int MAXN=200005;
set widthPoint,lengthPoint;
set::iterator leftPos,rightPos;
int widthSegment[MAXN],lengthSegment[MAXN];

void changeSegment(set &s,int *segmentArray,int newPoint)
{
    s.insert(newPoint);
    leftPos=rightPos=s.find(newPoint);
    leftPos--;
    rightPos++;
    segmentArray[*rightPos-*leftPos]--;
    segmentArray[newPoint-*leftPos]++;
    segmentArray[*rightPos-newPoint]++;
}
int main()
{
    memset(widthSegment,0,sizeof(widthSegment));
    memset(lengthSegment,0,sizeof(lengthSegment));
    int w,h,n,point,answ,ansl;
    char type[5];
    scanf("%d%d%d",&w,&h,&n);
    answ=w;
    ansl=h;
    widthPoint.insert(0);
    widthPoint.insert(w);
    lengthPoint.insert(0);
    lengthPoint.insert(h);
    widthSegment[w-0]=1;
    lengthSegment[h-0]=1;
    while(n--)
    {
        scanf("%s",type);
        scanf("%d",&point);
        if(type[0]=='V')
            changeSegment(widthPoint,widthSegment,point);
        else
            changeSegment(lengthPoint,lengthSegment,point);
        while(!widthSegment[answ])
            answ--;
        while(!lengthSegment[ansl])
            ansl--;
        printf("%I64d\n",(long long)answ*(long long)ansl);
    }
    return 0;
}