HDU 1541 树状数组

树状数组入门博客推荐 http://blog.csdn.net/qq_34374664/article/details/52787481

 Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9768    Accepted Submission(s): 3914

Problem Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 



For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. 

You are to write a program that will count the amounts of the stars of each level on a given map.

 

Input

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.

 

Output

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

 

Sample Input

5

1 1

5 1

7 1

3 3

5 5

Sample Output

1

2

1

1

0

 

题意  一共有n个点，已经按y坐标为第一优先级，x坐标为第二优先级排好序，输入。level的定义为 某个点的左下方存在点的个数（不包括本身）就是其level。问 0~n-1中 每个level点的个数。

思路  A[i]表示x坐标为i的个数 ( A[]无形数组)  C[]为A[]的树状数组，那么GetSum(i)就是 序列中前i个元素的和，即x小于等于i的点数。getsum(i)就是x坐标为i的点的level。

注意  0<=X,Y<=32000  我们要将x坐标都加1 才会避免函数add进入死循环

 

更加详细解读请参考博客 http://blog.csdn.net/ljd4305/article/details/10183285

 

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#define maxn 32050
using namespace std;
int c[maxn];
int level[maxn];
int lowbit(int x)
{
    return x&(-x);
}
int getsum(int x)
{
    int ans=0;
    while(x>0)
    {
        ans+=c[x];
        x-=lowbit(x);
    }
    return ans;
}
void add(int x,int y)
{
    while(x<maxn)
    {
        c[x]+=y;
        x+=lowbit(x);
    }
}
int main()
{
    int n;
    int x,y;
    int i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
    memset(c,0,sizeof(c));
    memset(level,0,sizeof(level));
    for(i=0;i<n;i++)
    {
        cin>>x>>y;
        x++;
        level[getsum(x)]++;
        add(x,1);
    }
    for(i=0;i<n;i++)
        cout<<level[i]<<endl;
    }
    return 0;
}