1/20集训一 STL F.（multiset 分割玻璃 ，求最大面积） Glass Carving

1/20集训一 STL

F.（multiset 分割玻璃 ，求最大面积） Glass Carving

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
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:

题意：
有一块w*h的玻璃， 现在可以水平或者垂直切割玻璃，给出切割步骤，求每一刀后的最大面积。

思路：
要求最大面积，就是要求最大的宽和高，我们可以将每次切割的位子用set保存（因为每次切割的位置是唯一的），然后每次切割就更新位置，同时将切割后得到的新的宽和高保存起来，记得要把被分割的这个原长给删掉。由于得到的宽和高是可以重复的，所以应该用multiset！！！一定不能用set！！！然后每次取multiset最后一个元素，即为最大值。

#include 
#include 
#include 
#include 
#include 

using namespace std;

int main()
{
    int w, h, n;
    cin>>w>>h>>n;
    set<int> qh, qv;
    multiset<int> mh, mv;
    qh.insert(h);
    qh.insert(0);
    qv.insert(w);
    qv.insert(0);
    mh.insert(h);
    mv.insert(w);
    char s[5];
    for(int i = 0; i < n; i++)
    {
        scanf("%s", s);
        set<int> ::iterator iter, pre, next;
        multiset<int> ::iterator era;
        if(s[0] == 'H')
        {
            int y;
            cin>>y;
            qh.insert(y);
            pre = next = iter = qh.find(y);
            pre--;
            next++;
            era = mh.find(*next - *pre);
            mh.erase(era);
            mh.insert(y - *pre);
            mh.insert(*next - y);
        }
        else if(s[0] == 'V')
        {
            int x;
            cin>>x;
            qv.insert(x);
            pre = next = iter = qv.find(x);
            pre--;
            next++;
            era = mv.find(*next - *pre);
            mv.erase(era);
            mv.insert(x - *pre);
            mv.insert(*next - x);
        }
        multiset<int> ::iterator iterh, iterv;
        iterh = mh.end();
        iterh--;
        iterv = mv.end();
        iterv--;
//      cout<<'*'<<*iterh<
//      cout<<'#'<<*iterv<
        long long ans = (long long)(*iterh) * (*iterv);
        printf("%lld\n", ans);
    }
    return 0;
}