HDU - 6183：Color it （线段树&动态开点||CDQ分治）

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.

0 0 : clear all the points.

1 1 x x y y c c : add a point which color is c c at point (x,y) (x,y) .

2 2 x x y 1  y1 y 2  y2 : 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 c c , that 1≤a≤x 1≤a≤x and y 1 ≤b≤y 2  y1≤b≤y2 , then the color c c should be counted.

3 3 : exit.

InputThe input contains many lines.

Each line contains a operation. It may be '0', '1 x y c' ( 1≤x,y≤10 6 ,0≤c≤50 1≤x,y≤106,0≤c≤50 ), '2 x y1 y2' (1≤x,y 1 ,y 2 ≤10 6  1≤x,y1,y2≤106 ) or '3'.

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

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

There are at most 150000 150000 continuous operations of operation 1 and operation 2.

There are at most 10 10 operation 0.

OutputFor each operation 2, output an integer means the answer .
Sample Input

0
1 1000000 1000000 50
1 1000000 999999 0
1 1000000 999999 0
1 1000000 1000000 49
2 1000000 1000000 1000000
2 1000000 1 1000000
0
1 1 1 1
2 1 1 2
1 1 2 2
2 1 1 2
1 2 2 2
2 1 1 2
1 2 1 3
2 2 1 2
2 10 1 2
2 10 2 2
0
1 1 1 1
2 1 1 1
1 1 2 1
2 1 1 2
1 2 2 1
2 1 1 2
1 2 1 1
2 2 1 2
2 10 1 2
2 10 2 2
3

Sample Output

2
3
1
2
2
3
3
1
1
1
1
1
1
1

题意：给定一个1e6*1e6的矩阵，有四种操作，0是清空矩阵；1是给某点加一个颜色c；2是查询某个区间(1,y1)到(x2,y2)的颜色数，3是退出。

思路：由于矩阵位置的特殊性，都是x=1开始，那么我们就是查询每种颜色在区间(y1,y2)是否有点的最小值小于x2。 用线段树维护区间最小值，动态开点。

（日后来补CDQ分治。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=2000010;
struct in{
    int l,r,v;
    in(){l=r=v=0;}
}s[maxn<<1]; int rt[51],cnt;
void update(int &Now,int L,int R,int pos,int val)
{
    if(!Now) Now=++cnt,s[Now].v=val;
    s[Now].v=min(s[Now].v,val);
    if(L==R) return ; int Mid=(L+R)>>1;
    if(pos<=Mid) update(s[Now].l,L,Mid,pos,val);
    else update(s[Now].r,Mid+1,R,pos,val);
}
bool query(int Now,int L,int R,int l,int r,int pos)
{
    if(!Now) return false;
    if(l<=L&&r>=R) return s[Now].v<=pos;
    int Mid=(L+R)>>1;
    if(l<=Mid) if(query(s[Now].l,L,Mid,l,r,pos)) return true;
    if(r>Mid) if(query(s[Now].r,Mid+1,R,l,r,pos)) return true;
    return false;
}
int main()
{
    int opt,x1,y1,x2,y2,c;
    while(scanf("%d",&opt)){
        if(opt==0){
            rep(i,0,50) rt[i]=0;
            rep(i,1,cnt) s[i].l=s[i].r=s[i].v=0;cnt=0;
        }
        else if(opt==1){
            scanf("%d%d%d",&x1,&y1,&c);
            update(rt[c],1,1000000,y1,x1);
        }
        else if(opt==2){
            scanf("%d%d%d",&x2,&y1,&y2); int ans=0;
            rep(i,0,50) ans+=query(rt[i],1,1000000,y1,y2,x2);
            printf("%d\n",ans);
        }
        else break;
    }
    return 0;
}