2018.2.1【 CodeForces - 527C 】解题报告（STL，set，边长和位置的维护）

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.

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:

Picture for the second sample test:

【题目大意】

看图就能理解，给一块儿长款确定的玻璃，每一次在长/宽处切一刀，问多次操作后分成的小玻璃最大面积是多少。

【解题思路】

利用set容器，维护位置，用数组来记录所有的边长，由于一定递减，所以每一次操作以后向下寻找最大面积。

【解题代码】

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=2e5+5;
typedef long long ll;
set::iterator i, j;  //迭代器
set ve, ho;  //记录所有边的位置 
int  wi[maxn],hi[maxn];

void cut(set &s,int *worh,int p)	//主要的维护函数 
{
	s.insert(p), i=j=s.find(p);//先插入一个长，宽的值，在set里寻找到p的位置（有序） 
	--i,++j,--worh[*j-*i];	//记录p左右相邻大小的长/宽，然后把相邻的这个长/宽位置差，即这个边长从set里删掉。
	++worh[p-*i],++worh[*j-p];	//新加入两块子边 
}

int main()  
{  
    int w, n, h, p, mw, mh;  
    char s[10];  
    while(~scanf("%d%d%d", &w, &h, &n))  
    {  
        memset(wi, 0, sizeof(wi)), memset(hi, 0, sizeof(hi));  	//边长数组清零 
        ve.clear(), ho.clear();  	//set容器清空初始化,加入0和最大值 
        ve.insert(0), ho.insert(0);  
        ve.insert(w), ho.insert(h);  
        wi[w] = hi[h] = 1;  
        mw = w , mh = h;  //记录初始最大值 
        while(n--)  
        {  
            scanf("%s%d", s, &p);  
            if(s[0] == 'V') cut(ve, wi, p);  
            else cut(ho, hi, p);  
            while(!wi[mw]) --mw;  	//向下寻找最大值 
            while(!hi[mh]) --mh;  
            printf("%lld\n", (ll)mw*(ll)mh);  
        }  
    }  
    return 0;  
}  

【收获与反思】

主要参考学习了这位前辈的代码：点击打开链接。

1.利用std::set::find方法找到插入的位置，从左右选取临近的值来计算新的长度。

2.利用单调性，以边长的值为index向下寻找，数组其实时bool形式，1为有，2为无。