hdu 6183 Color it （线段树 动态开点）

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.

InputThe 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.

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

思路：有50种颜色，对每一种颜色建一颗线段树维护，动态开点。第一种操作：使点（x,y）的颜色变为c第二种：询问（1，y1),(x,y2)两点间的颜色种类数量我们可以以y轴建线段树，横坐标为值，那么要确定两点之前是否存在某种颜色，只要询问下每个颜色在y1,y2之间最小的值(也就是横坐标）.判断下最小值是否小于第二种操作给出的x,如果小于的话就代表两点间存在这种颜色。

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mid int m = (l + r) >> 1
const int M = 1e6+10;
int n = 1e6,flag=0;
int sum[M],ne=0,ls[M],rs[M],rt[100];

void update(int &k,int l,int r,int p,int num){
    if(!k){   //如果当前区间未拓展，拓展并赋值
        k = ++ne;
        sum[k] = num;
    }
    sum[k] = min(sum[k],num);//当前区间有值，更新下最小值
    if(l == r)
        return ;
    mid;
    if(p <= m) update(ls[k],l,m,p,num);
    else update(rs[k],m+1,r,p,num);
}

void query(int k,int L,int R,int l,int r,int up){
    if(!k||flag) return ;
    if(L <= l&&R >= r){
        if(sum[k]<=up)
            flag = 1;
            return;
    }
    mid;
    if(L <= m) query(ls[k],L,R,l,m,up);
    if(R > m) query(rs[k],L,R,m+1,r,up);
    return ;
}

void  init(){
   memset(rt,0,sizeof(rt));
   memset(sum,0,sizeof(sum));
   memset(ls,0,sizeof(ls));
   memset(rs,0,sizeof(rs));
   ne = 0;
}
int main()
{
    int op,x,y,z;
    while(~scanf("%d",&op)){
        if(op == 0)
            init();
        else if(op == 1){
            scanf("%d%d%d",&x,&y,&z);
            update(rt[z],1,n,y,x);
        }
        else if(op == 2){
            scanf("%d%d%d",&x,&y,&z);
            int ans = 0 ;
            for(int i = 0;i <= 50;i ++){
                 flag = 0;
                 query(rt[i],y,z,1,n,x);
                 if(flag) ans++;
            }
            printf("%d\n",ans);
        }
        else return 0;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/kls123/p/8992605.html