2017ACM/ICPC广西邀请赛 Color it

Color it

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

 

题解：这道题目AC的人比较少    在这提供俩种想法

第一：题目规定了颜色的数目最多是51种    

我们可以使用51个线段树   来保存每一种颜色的那些坐标

然后使用线段树查询     时间可能会少点    

第二：我是使用的第二种方法，题目说了最多有150000个1和2的操作

所以穷举也不是不可以的考虑的    我就是使用穷举AC的

不过时间复杂度不能和线段树的比

//但是   我的代码简单    
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <cstring>
#include <vector>
#include <math.h>
using namespace std;

struct Node
{
    int x;
    int y;
} node;
vector<Node>v[65];
int main()
{
    int k,c;
    int x2,y1,y2;
    while(1)
    {
        scanf("%d",&k);
        if(k==3)break;
        if(k==0)
        {
            for(int i=0; i<65; ++i)
                v[i].clear();
        }
        else if(k==1)
        {
            scanf("%d%d%d",&node.x,&node.y,&c);
            v[c].push_back(node);
        }
        else
        {
            int ans=0;
            scanf("%d%d%d",&x2,&y1,&y2);
            for(int i=0; i<=50; ++i)
            {
                for(int j=0; j<v[i].size(); ++j)
                {
                    int xx=v[i][j].x;
                    int yy=v[i][j].y;
                    if(xx<=x2&&yy<=y2&&yy>=y1)
                    {
                        ans++;
                        break;
                    }
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}
//欢迎喜欢算法  IT的dalao加1345411028  带我飞