Stars（二维树状数组）

Stars

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)

Total Submission(s): 111 Accepted Submission(s): 54

Problem Description

Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y" where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.

There is only one case.

 

Input

The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.

 

Output

For each query,output the number of bright stars in one line.

 

Sample Input

5
B 581 145
B 581 145
Q 0 600 0 200
D 581 145
Q 0 600 0 200

 

Sample Output

1
0

 

Author

teddy

 

Source

百万秦关终属楚

 

Recommend

teddy

 

/*
题意：二维坐标，然后三种操作，B x y 坐标(x,y)处的星星亮了，D x y坐标(x,y)处的星星灭了 Q X1 X2 Y1 Y2询问矩形内有多少颗
    亮的星星
    
初步思路：二维树状数组，但是操作的时候先要查询一下当前位置星星的状态,还有就是查询的坐标可能不是x1<x2 y1<y2
*/
#include<bits/stdc++.h>
#define N 1010
#define lowbit(x) x&(-x)
using namespace std;
int t,X1,X2,Y1,Y2;
int n1,n2,m1,m2;
char op[2];
int c[N][N];
void update(int x,int y,int val)
{
    for(int i=x;i<N;i+=lowbit(i))
    {
        for(int j=y;j<N;j+=lowbit(j))
        {
            c[i][j]+=val;  
        }    
    }
    // return val;//返回你实际搬运的东西
}
int getsum(int x,int y)
{
    int s=0;
    for(int i=x;i>0;i-=lowbit(i))
    {
        for(int j=y;j>0;j-=lowbit(j))
        {
            s+=c[i][j];
        }
    }
    return s;
}
void init(){
    memset(c,0,sizeof c);
}
int main(){
    // freopen("in.txt","r",stdin);
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%s",op);
        if(op[0]=='B'){
            scanf("%d%d",&X1,&Y1);
            X1++;
            Y1++;
            //先查询一下当前位置的星星是不是亮着的
            if(getsum(X1,Y1)-getsum(X1-1,Y1)-getsum(X1,Y1-1)+getsum(X1-1,Y1-1)==1){
                // cout<<"有星星亮的"<<endl;
                continue;
            }
            
            update(X1,Y1,1);
            
        }else if(op[0]=='D'){
            scanf("%d%d",&X1,&Y1);
            X1++;
            Y1++;
            //先查询一下当前位置的星星是不是没亮
            if(getsum(X1,Y1)-getsum(X1-1,Y1)-getsum(X1,Y1-1)+getsum(X1-1,Y1-1)==0){
                // cout<<"星星已经灭了"<<endl;
                continue;
            }
            update(X1,Y1,-1);
            
        }else{
            scanf("%d%d%d%d",&X1,&X2,&Y1,&Y2);
            X1++;X2++;
            Y1++;Y2++;
            // cout<<X1<<" "<<Y1<<" "<<X2<<" "<<Y2<<endl;
            // cout<<getsum(X2,Y2)<<" "<<getsum(X1-1,Y2)<<" "<<getsum(X2,Y1-1)<<" "<<getsum(X1-1,Y1-1)<<endl;
            n1=min(X1,X2),m1=min(Y1,Y2),n2=max(X1,X2),m2=max(Y1,Y2);
            printf("%d\n",getsum(n2,m2)-getsum(n1-1,m2)-getsum(n2,m1-1)+getsum(n1-1,m1-1));
        }
    }
    return 0;
}