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?
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.
After each cut print on a single line the area of the maximum available glass fragment in mm2.
4 3 4 H 2 V 2 V 3 V 1
8 4 4 2
7 6 5 H 4 V 3 V 5 H 2 V 1
28 16 12 6 4
Picture for the first sample test:
题意：一个w*h的玻璃，如今水平或竖直切n次（“H”表示水平切。“V”表示竖直切），每一次切后输出当前切成的块中的最大面积。
思路：用set记录分割的位置（要用两个set，分别来记录长和宽），multiset记录某一条边被切后 所得到的 小段的长度（也要两个，分别记录长和宽的）。
那么每次切后就从multiset中取出最大的长和宽。相乘即得面积。
STL set 写法
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi
")
typedef __int64 ll;
using namespace std;
int w,h,n;
set<int>wxs;
set<int>hxs;
multiset<int>wds;
multiset<int>hds;
int main()
{
int i,j;
while (~sfff(w,h,n))
{
set<int>::iterator it,p;
char s[3];
int x;
wxs.clear();
hxs.clear();
wds.clear();
hds.clear();
wxs.insert(0); wxs.insert(w);
hxs.insert(0); hxs.insert(h);
wds.insert(w); hds.insert(h);
while (n--)
{
scanf("%s%d",s,&x);
if (s[0]=='H')
{
it=hxs.lower_bound(x);
p=it;
p--;
int dis = *it - *p;
hds.erase(hds.find(dis));//这里不能写成hds.erase(dis),在multiset里面这样写会把全部值等于dis的点删掉，这显然不符合我们的题意
hds.insert(*it-x);
hds.insert(x-*p);
hxs.insert(x);
}
else
{
it=wxs.lower_bound(x);
p=it;
p--;
int dis = *it - *p;
wds.erase(wds.find(dis));
wds.insert(*it-x);
wds.insert(x-*p);
wxs.insert(x);
}
int xx= *wds.rbegin();
int yy= *hds.rbegin();
pf("%I64d
",(ll)xx * (ll)yy); //最后要强制转化，不然会爆int
}
}
return 0;
}
并查集写法
思路：并查集初始化。首先将玻璃所有切成1*1的小块，然后先保存下所有的操作。记录下它切了哪些位置（数组vis_w和vis_h）。接着将没有被切的位置 i 连起来Union（i，i+1），最后倒着把要切的位置连起来，这个过程中记录每次两条边的最大值，它们的乘积保存下来就是答案。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 200010
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi
")
typedef __int64 ll;
using namespace std;
//维护两个并查集。分别维护w和h
int w,h,n,max_w,max_h;
int father_w[maxn],father_h[maxn];
int num_w[maxn],num_h[maxn];
bool vis_w[maxn],vis_h[maxn];
char s[maxn][3];
int pos[maxn];
ll ans[maxn];
void init()
{
int i;
FRL(i,0,maxn)
{
father_w[i]=i;
father_h[i]=i;
num_w[i]=1;
num_h[i]=1;
}
num_w[0]=0;
num_h[0]=0;
mem(vis_w,false);
mem(vis_h,false);
max_w=1;//用来记录每次Union操作后边的最大值
max_h=1;
}
int find_father_w(int x)
{
if (x!=father_w[x])
father_w[x]=find_father_w(father_w[x]);
return father_w[x];
}
int find_father_h(int x)
{
if (x!=father_h[x])
father_h[x]=find_father_h(father_h[x]);
return father_h[x];
}
void Union_w(int a,int b)
{
int fa=find_father_w(a);
int fb=find_father_w(b);
if (fa!=fb)
{
father_w[fb]=fa;
num_w[fa]+=num_w[fb];
}
max_w=max(max_w,num_w[fa]);
}
void Union_h(int a,int b)
{
int fa=find_father_h(a);
int fb=find_father_h(b);
if (fa!=fb)
{
father_h[fb]=fa;
num_h[fa]+=num_h[fb];
}
max_h=max(max_h,num_h[fa]);
}
int main()
{
int i,j;
while (~sfff(w,h,n))
{
init();
FRE(i,1,n)
{
scanf("%s%d",s[i],&pos[i]);
if (s[i][0]=='H') vis_h[pos[i]]=true;
else vis_w[pos[i]]=true;
}
FRL(i,1,w)//先把没有被切的位置连起来
{
if (!vis_w[i])
Union_w(i,i+1);
}
FRL(i,1,h)
{
if (!vis_h[i])
Union_h(i,i+1);
}
for (i=n;i>0;i--)//逆操作连接
{
// pf("max_w=%d
max_h=%d
***
",max_w,max_h);
ans[i]=(ll)max_w * (ll)max_h;//保存答案
if (s[i][0]=='H') Union_h(pos[i],pos[i]+1);
else Union_w(pos[i],pos[i]+1);
}
FRE(i,1,n)
pf("%I64d
",ans[i]);
}
return 0;
}