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[];
int main()
{
    int k,c;
    int x2,y1,y2;
    while()
    {
        scanf("%d",&k);
        if(k==)break;
        if(k==)
        {
            for(int i=; i<; ++i)
                v[i].clear();
        }
        else if(k==)
        {
            scanf("%d%d%d",&node.x,&node.y,&c);
            v[c].push_back(node);
        }
        else
        {
            int ans=;
            scanf("%d%d%d",&x2,&y1,&y2);
            for(int i=; i<=; ++i)
            {
                for(int j=; 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 ;
}
//欢迎喜欢算法  IT的dalao加1345411028  带我飞