CF #296 (Div. 1) A. Glass Carving 线段树

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 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:


题目意思：
一个w*h的矩形，然后用水平线和竖直线切割，每次切割后输出最大的子面积。

思路：
很明显最大子面积是横着最大的长度*竖着最大的长度，那么建两棵线段树，每个区间lx、rx、maxh分别为离该区间左端点最近的已经切割过的点、离右端点最近的已经切割过的点、最长的切割长度。维护这三个值就行了。代码略丑。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 200005
#define ll root<<1
#define rr root<<1|1
#define mid1 (a[root].l+a[root].r)/2
#define mid2 (b[root].l+b[root].r)/2

int w, h, n;
struct node{
int l, r;
int lx, rx;
int maxh;
}a[N*4], b[N*4];

void build1(int l,int r,int root){
a[root].l=l;
a[root].r=r;
if(l==r){
if(l==0) a[root].lx=a[root].rx=0;
else if(l==w) a[root].lx=a[root].rx=w;
else a[root].lx=a[root].rx=-1;
a[root].maxh=0;
return;
}
build1(l,mid1,ll);
build1(mid1+1,r,rr);
a[root].lx=a[ll].lx;
a[root].rx=a[rr].rx;
a[root].maxh=a[root].r-a[root].l;
}

void build2(int l,int r,int root){
b[root].l=l;
b[root].r=r;
if(l==r){
if(l==0) b[root].lx=b[root].rx=0;
else if(l==h) b[root].lx=b[root].rx=h;
else b[root].lx=b[root].rx=-1;
b[root].maxh=0;
return;
}
build2(l,mid2,ll);
build2(mid2+1,r,rr);
b[root].lx=b[ll].lx;
b[root].rx=b[rr].rx;
b[root].maxh=b[root].r-b[root].l;
}

void update1(int p,int root){
if(a[root].l==p&&a[root].r==p){
a[root].lx=a[root].rx=p;return;
}
if(p<=a[ll].r) update1(p,ll);
else update1(p,rr);
int ln, rn;
if(a[ll].lx!=-1) a[root].lx=a[ll].lx;
else if(a[ll].rx!=-1) a[root].lx=a[ll].rx;
else if(a[rr].lx!=-1) a[root].lx=a[rr].lx;
else if(a[rr].rx!=-1) a[root].lx=a[rr].rx;
else a[root].lx=-1;

if(a[rr].rx!=-1) a[root].rx=a[rr].rx;
else if(a[rr].lx!=-1) a[root].rx=a[rr].lx;
else if(a[ll].rx!=-1) a[root].rx=a[ll].rx;
else if(a[ll].lx!=-1) a[root].rx=a[ll].lx;
else a[root].rx=-1;

if(a[ll].rx!=-1) ln=a[ll].r-a[ll].rx;
else ln=a[ll].r-a[ll].l;
if(a[rr].lx!=-1) rn=a[rr].lx-a[rr].l;
else rn=a[rr].r-a[rr].l;

a[root].maxh=max(max(a[ll].maxh,a[rr].maxh),ln+rn+1);
}

void update2(int p,int root){
if(b[root].l==p&&b[root].r==p){
b[root].lx=b[root].rx=p;return;
}
if(p<=b[ll].r) update2(p,ll);
else update2(p,rr);
int ln, rn;
if(b[ll].lx!=-1) b[root].lx=b[ll].lx;
else if(b[ll].rx!=-1) b[root].lx=b[ll].rx;
else if(b[rr].lx!=-1) b[root].lx=b[rr].lx;
else if(b[rr].rx!=-1) b[root].lx=b[rr].rx;
else b[root].lx=-1;

if(b[rr].rx!=-1) b[root].rx=b[rr].rx;
else if(b[rr].lx!=-1) b[root].rx=b[rr].lx;
else if(b[ll].rx!=-1) b[root].rx=b[ll].rx;
else if(b[ll].lx!=-1) b[root].rx=b[ll].lx;
else b[root].rx=-1;
if(b[ll].rx!=-1) ln=b[ll].r-b[ll].rx;
else ln=b[ll].r-b[ll].l;
if(b[rr].lx!=-1) rn=b[rr].lx-b[rr].l;
else rn=b[rr].r-b[rr].l;
b[root].maxh=max(max(b[ll].maxh,b[rr].maxh),ln+rn+1);
}

main()
{
int i, j, k;
while(scanf("%d %d %d",&w,&h,&n)==3){
char s[5];
build1(0,w,1);
build2(0,h,1);
//    printf("%d %d\n",a[1].maxh,b[1].maxh);
while(n--){
scanf("%s%d",s,&k);
if(s[0]=='H'){
update2(k,1);
}
else{
update1(k,1);
}
//    printf("%d %d\n",a[1].maxh,b[1].maxh);
printf("%I64d\n",(__int64)a[1].maxh*(__int64)b[1].maxh);
}
}
}