CF # 296 C Glass Carving （并查集 或者 multiset）

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:



题意:切玻璃，每切一刀求当前的最大快的面积

思路: 最重要到额一点是找横的最大和竖着的最大

如果用 set,每次把新的长宽插入，然后erase（）被破坏的长或者宽

如果用并查集，先把没有切的看成一个整体，然后重后往前依次去掉一条切得线后可以得到新的一个长或者宽

multiset 代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#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 bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 10005

multiset<ll>x,y;
multiset<ll>hx,hy;
multiset<ll>::iterator it;

char ch[10];

ll w,h;
ll xx;
int n;

int main()
{
    int i,j;
    scanf("%I64d%I64d%d",&w,&h,&n);

	x.insert(0);
	x.insert(w);

	y.insert(0);
	y.insert(h);

    hx.insert(w);
    hy.insert(h);

    while(n--)
	{
       scanf("%s%d",ch,&xx);
       if(ch[0]=='V')
	   {
	   	  x.insert(xx);

	   	  it=x.find(xx);
	   	  it++;
	   	  ll ri=*it;
	   	  it--;
	   	  it--;
	   	  ll le=*it;
          it=hx.find(ri-le);  //这一行和下一行不能用hx.erase(ri-le)代替，因为这样是删除所有的
	   	  hx.erase(it);       //插入一条线，破坏了一个长度
	   	  hx.insert(ri-xx);   //增加了两个长度
	   	  hx.insert(xx-le);
	   }
       else
	   {
	   	  y.insert(xx);

	   	  it=y.find(xx);
	   	  it++;
	   	  ll ri=*it;
	   	  it--;
	   	  it--;
	   	  ll le=*it;

          it=hy.find(ri-le);
	   	  hy.erase(it);
	   	  hy.insert(ri-xx);
	   	  hy.insert(xx-le);
	   }

	   it=hx.end();
	   it--;
	   ll le=*it;

	   it=hy.end();      //长宽都取最大值
	   it--;
	   ll ri=*it;

	   pf("%I64d\n",le*ri);
	}
    return 0;
}

并查集代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#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 bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 200005

int lenh,lenw;    //长的最大值与宽的最大值
int fah
,faw
;
int numh
,numw
;
int vish
,visw
;
int w,h,n;
int op
;
ll ans
;
char ch
[3];

int chah(int x)
{
   if(fah[x]!=x)
	fah[x]=chah(fah[x]);
   return fah[x];
}

void unith(int x,int y)
{
    x=chah(x);
    y=chah(y);

   if(x==y) return ;

   fah[y]=x;
   numh[x]+=numh[y];
   lenh=max(numh[x],lenh);
}

int chaw(int x)
{
   if(faw[x]!=x)
	faw[x]=chaw(faw[x]);
   return faw[x];
}

void unitw(int x,int y)
{
    x=chaw(x);
    y=chaw(y);

   if(x==y) return ;

   faw[y]=x;
   numw[x]+=numw[y];

   lenw=max(numw[x],lenw);
}

int main()
{
    int i,j;
    sfff(w,h,n);

    mem(visw,0);
    mem(vish,0);
    int t=max(h,w);

    fre(i,0,t+1)
    {
        fah[i]=i;
        numh[i]=1;
        faw[i]=i;
        numw[i]=1;
    }

    numh[0]=numw[0]=0;

    int x;

    fre(i,0,n)
    {
       scanf("%s%d",ch[i],&op[i]);

       if(ch[i][0]=='H')
         vish[op[i]]=1;
       else
         	visw[op[i]]=1;

    }

    lenh=1;
    lenw=1;

    fre(i,1,h)
        if(!vish[i])
            unith(i,i+1);

    fre(i,1,w)
       if(!visw[i])
			unitw(i,i+1);

//    fre(i,1,h)
//      if(fah[i]==i)
//	  {
//	  	pf("%d %d\n",i,numh[i]);
//
//	  }

//    fre(i,1,w)
//      if(faw[i]==i)
//	  {
//	  	pf("%d %d\n",i,numw[i]);
//
//	  }

    free(i,n-1,0)
    {
        //pf("****%d %d\n",lenw,lenh);

    	ans[i]=(ll)(lenh)*(ll)(lenw);

    	if(ch[i][0]=='H')
			unith(op[i],op[i]+1);
		else
			unitw(op[i],op[i]+1);
    }

   fre(i,0,n)
    {
    	pf("%I64d\n",ans[i]);
    }

  return 0;
}