***hdu6183

轉載自：http://blog.csdn.net/oranges_c/article/details/77800825

Do you like painting? Little D doesn’t like painting, especially messy color paintings. Now Little B is painting. To prevent him from drawing messy painting, Little D asks you to write a program to maintain following operations. The specific format of these operations is as follows.

00 : clear all the points.

11 xx yy cc : add a point which color is cc at point (x,y)(x,y).

22 xx y1y1 y2y2 : count how many different colors in the square (1,y1)(1,y1) and (x,y2)(x,y2). That is to say, if there is a point (a,b)(a,b) colored cc, that 1≤a≤x1≤a≤x and y1≤b≤y2y1≤b≤y2, then the color cc should be counted.

33 : exit.
Input
The input contains many lines.

Each line contains a operation. It may be ‘0’, ‘1 x y c’ ( 1≤x,y≤106,0≤c≤501≤x,y≤106,0≤c≤50 ), ‘2 x y1 y2’ (1≤x,y1,y2≤1061≤x,y1,y2≤106 ) or ‘3’.

x,y,c,y1,y2x,y,c,y1,y2 are all integers.

Assume the last operation is 3 and it appears only once.

There are at most 150000150000 continuous operations of operation 1 and operation 2.

There are at most 1010 operation 0.

Output
For each operation 2, output an integer means the answer .

思路：
1，由於顏色只有51種所以我們可以分別建立51棵線段樹，每一棵線段樹維護一種顏色值；
2，由於查詢的時候橫座標的較小值一直都是一，變化的只是較大的那個橫座標，縱座標兩個都在變。所以，我們可以將縱座標作爲線段樹的區間，然後在每個區間內維護這個區間中的出現的最小橫座標值，如果連最小的橫座標都比查詢的較大的那個橫座標大，那麼就說明這個矩形區域中沒有我們要找的點；

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int tot=0,T[59],v[maxn],ls[maxn],rs[maxn],flag=0;
int op,x,y,c,y2;
void update(int &t,int y,int x,int l,int r)
{
    if(!t)
    {
        t=++tot;
        v[t]=x;
    }
    v[t]=min(v[t],x);
    if(l==r) return;
    int m=l+r>>1;
    if(y<=m)
        update(ls[t],y,x,l,m);
    else update(rs[t],y,x,m+1,r);
}
void query(int t,int L,int R,int x,int l,int r)
{
    if(flag||!t)
        return;
   if(L<=l&&r<=R)
   {
       if(v[t]<=x)
        flag=1;
       return;
   }
   int m=l+r>>1;
   if(L<=m) query(ls[t],L,R,x,l,m);
   if(m<R) query(rs[t],L,R,x,m+1,r);
}
int main()
{
    while(~scanf("%d",&op))
    {
        if(op==0)
        {
           memset(ls,0,sizeof ls);
           memset(rs,0,sizeof rs);
           memset(T,0,sizeof T);
           tot=0;
        }
        else if(op==1)
        {
            scanf("%d%d%d",&x,&y,&c);
            update(T[c],y,x,1,1e6);
        }
        else if(op==2)
        {
            scanf("%d%d%d",&x,&y,&y2);
            int ans=0;
            for(int i=0;i<=50;i++)
            {
                flag=0;
                query(T[i],y,y2,x,1,1e6);
                ans+=flag;
            }
            printf("%d\n",ans);
        }
        else return 0;
    }
    return 0;
}