Codeforces Round #296 (Div. 1) A. Glass Carving Set的妙用

A. 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.

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：

给你n*m大的矩形，然后会切k刀，让你输出每切一刀后所分割的小矩形的最大面积是多少

题解

最大的小矩形，那么就应该是最大的长乘以最大的宽，我们用一个set存储所有切的位置，用一个数列来存储切之后每一段的长度是多少

然后我们一次递减查询就好，具体看代码吧~

代码：

int lenx[],leny[];
set<int> hor,ver;
int main()
{
    int w,h,n,dis;
    scanf("%d%d%d",&w,&h,&n);
    hor.insert();
    hor.insert(h);
    leny[]++;
    leny[h]++;
    ver.insert();
    ver.insert(w);
    lenx[]++;
    lenx[w]++;
    set<int>::iterator itr,itr1,itr2;
    char op[];
    int maxy=h,maxx=w;
    REP(i,n)
    {
        scanf("%s %d",op,&dis);
        if(op[]=='H')
        {
            hor.insert(dis);
            itr=hor.find(dis);
            itr1=itr;
            itr2=itr;
            itr1--;
            itr2++;
            leny[*itr2-*itr1]--;
            leny[*itr2-*itr]++;
            leny[*itr-*itr1]++;
        }
        else
        {
            ver.insert(dis);
            itr=ver.find(dis);
            itr1=itr;
            itr2=itr;
            itr1--;
            itr2++;
            lenx[*itr2-*itr1]--;
            lenx[*itr2-*itr]++;
            lenx[*itr-*itr1]++;
        }
        while(!leny[maxy])maxy--;
        while(!lenx[maxx])maxx--;
        //cout<<maxy<<" "<<maxx<<endl;
        printf("%I64d\n",1LL*maxx*maxy);
    }
}