Points in Rectangle (树状数组）

Description

As the name says, this problem is about finding the number of points in a rectangle whose sides are parallel to axis. All the points and rectangles consist of 2D Cartesian co-ordinates. A point that lies in the boundary of a rectangle is considered inside.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer q (1 ≤ q ≤ 30000) denoting the number of queries. Each query is either one of the following:

1)      0 x y, meaning that you have got a new point whose co-ordinate is (x, y). But the restriction is that, if a point (x, y) is already listed, then this query has no effect.

2)      1 x₁ y₁ x₂ y₂ meaning that you are given a rectangle whose lower left co-ordinate is (x₁, y₁) and upper-right corner is (x₂, y₂); your task is to find the number of points, given so far, that lie inside this rectangle. You can assume that (x₁ < x₂, y₁ < y₂).

You can assume that the values of the co-ordinates lie between 0 and 1000 (inclusive).

Output

For each case, print the case number in a line first. Then for each query type (2), you have to answer the number of points that lie inside that rectangle. Print each of the results in separated lines.

Sample Input

1

9

0 1 1

0 2 6

1 1 1 6 6

1 2 2 5 5

0 5 5

1 0 0 6 5

0 3 3

0 2 6

1 2 1 10 10

Sample Output

Case 1:

2

0

2

3

题意：很多次操作

1.在平面上选一点

2.选择一个与x，y轴平行的矩形，问矩形中含多少点。

典型的二维树状数组

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int tree[1005][1005],N;bool mark[1005][1005];void update(int x,int y,int change){    while(x<=1004)    {        int y1=y;        while(y1<=1004)        {            tree[x][y1]+=change;            y1+=y1&(y1^(y1-1));        }        x+=x&(x^(x-1));    }}int query(int x,int y){    int coun=0;    while(x>0)    {        int y2=y;        while(y2>0)        {            coun+=tree[x][y2];            y2-=y2&(y2^(y2-1));        }        x-=x&(x^(x-1));    }    return coun;}int main(){     int times;    scanf("%d",×);    for(int time=1;time<=times;++time)    {     memset(tree,0,sizeof(tree));     memset(mark,0,sizeof(mark));    int q;    printf("Case %d:\n",time);    scanf("%d",&q);    for(int i=1;i<=q;++i)    {        int que;        scanf("%d",&que);        if(que)        {            int x1,y1,x2,y2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);           // int c22=query(x2,y2);           printf("%d\n",query(x2+1,y2+1)-query(x2+1,y1)-query(x1,y2+1)+query(x1,y1));        }        else        {            int x,y;            scanf("%d%d",&x,&y);            if(mark[x+1][y+1])                continue;            else            {                mark[x+1][y+1]=true;                update(x+1,y+1,1);            }        }    }    }    return 0;}

其中有两点需要注意

1.树状数组的下标应当大于0，若题中含0，或负数，可以加上最小负数的绝对值+1，全部转为正数。

2.常常涉及到容斥定理，对于其中的边界取舍，应当注意最好画个草图，以区分界限，保证只统计一次。