hdu6183 Color it 线段树动态开点+查询减枝

题目传送门

题目大意：

　　有多次操作。操作0是清空二维平面的点，操作1是往二维平面（x,y）上放一个颜色为c的点，操作2是查询一个贴着y轴的矩形内有几种颜色的点，操作3退出程序。

思路：

　　由于查询的矩形是贴着y轴的，所以以y轴为线段树节点，建立52颗线段树，然后每个节点都保存这个纵坐标下x的最小值，然后查询。

　　这样的线段树显然是开不下的，所以我们考虑线段树动态开点，但是发现有50颗，如果按照50*n*logn的查询，还是会TLE，这里需要一个减枝，就是如果一部分区间内已经有一个颜色了，就直接退出所有查询即可，否则会TLE（卡常数？）

#include<bits/stdc++.h>
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=100005;
int rt[52],tot;
int R[maxn*20],L[maxn*20],v[maxn*20],flag;
int op,x,y,y1,y2,co;
void init(){
    tot=0,clr(rt,0);
}
void update(int &o,int l,int r,int y,int x){
    if(!o){
        L[o=++tot]=0,R[o]=0;
        v[o]=x;
    }
    v[o]=min(v[o],x);
    if(l==r)return;
    int mid=(l+r)>>1;
    if(y<=mid)update(L[o],l,mid,y,x);
    else update(R[o],mid+1,r,y,x);
}
void query(int o,int l,int r,int ql,int qr){
    if(flag||!o){
        return ;
    }
    if(ql<=l&&qr>=r)
    {
     if(v[o]<=x)flag=1;
     return;
    }
    int mid=(l+r)>>1;
    
    if(ql<=mid)query(L[o],l,mid,ql,qr);
    if(qr>mid)query(R[o],mid+1,r,ql,qr);
    return ;
}
int main(){
//    freopen("simple.in","r",stdin);
    int n=1e6;
    while(scanf("%d",&op)!=EOF){
        if(op==3)break;
        else if(op==0){
            init();
        }else if(op==1){
            scanf("%d%d%d",&x,&y,&co);
            update(rt[co],1,n,y,x);
        }else{
            scanf("%d%d%d",&x,&y1,&y2);
            int ans=0;
            for(int i=0;i<=50;i++)
            {
                flag=0;
                query(rt[i],1,n,y1,y2);
                ans+=flag;
            }
            printf("%d
",ans);
        }
    }
}

Color it

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 2327    Accepted Submission(s): 703

Problem Description

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 : clear all the points.

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

2 x y1 y2 : count how many different colors in the square (1,y1) and (x,y2). That is to say, if there is a point (a,b) colored c, that 1≤a≤x and y1≤b≤y2, then the color c should be counted.

3 : 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≤50 ), '2 x y1 y2' (1≤x,y1,y2≤106 ) or '3'. 

x,y,c,y1,y2 are all integers.

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

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

There are at most 10 operation 0. 

 

Output

For 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

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

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