Codeforces 527C Glass Carving<set集合和multiset集合的使用>

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:



 解题思路：用2个set分别记录长和宽的切割点，再用两个multiset记录切割后的长和宽中区间段的长度。

初始将长宽边界和长度插入set和multiset，再将切割点插入set，用find找出其位置，确定相邻切割点(a,b)，在相应的multiset中删除一个(b-a)区间，最大的面积就是两个multiset中最大值的乘积。

利用set与multiset能够简化我们排序，查询，更新各种操作，所以是很好的集合使用的题目，
muitiset与set的区别在于前者可以出现重复的值，后者不允许出现重复值。

#include<cstdio>
#include<algorithm>
#include<set>
#include<iostream>
using namespace std;
set<int> h;
set<int> w;
multiset<int>his;
multiset<int>wis;
int main (){
int wi,hi,n;
while(~scanf("%d%d%d",&wi,&hi,&n)){
h.clear();
w.clear();
his.clear();
wis.clear();
h.insert(hi);
w.insert(wi);
h.insert(0);
w.insert(0);
his.insert(hi);
wis.insert(wi);
char o;
int len;
while(n--){
getchar();
scanf("%c%d",&o,&len);
//cout<<o<<" "<<len<<endl;
if(o=='H'){
h.insert(len);
int a=*(--h.find(len));
int b=*(++h.find(len));
his.erase(his.find(b-a));
his.insert(len-a);
his.insert(b-len);
}
else
{
w.insert(len);
int a=*(--w.find(len));
int b=*(++w.find(len));
wis.erase(wis.find(b-a));
wis.insert(len-a);
wis.insert(b-len);
}
long long high=*(--his.end());
long long weigh=*(--wis.end());
// cout<<high<<" "<<weigh<<endl;
printf("%lld\n",high*weigh);
}
}

}