***hdu6183

转载自：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，由于颜色只有51种所以我们可以分别建立51棵线段树，每一棵线段树维护一种颜色值；
2，由于查询的时候横坐标的较小值一直都是一，变化的只是较大的那个横坐标，纵坐标两个都在变。所以，我们可以将纵坐标作为线段树的区间，然后在每个区间内维护这个区间中的出现的最小横坐标值，如果连最小的横坐标都比查询的较大的那个横坐标大，那么就说明这个矩形区域中没有我们要找的点；

#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;}