[ACM] hdu 2352 Stars （树状数组）

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30272		Accepted: 13206

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

解题思路：

题意：按y递增，如果y相同按x递增的顺序给出n个点，如果某个点的坐标为（x0，y0), 它的左下角（包括正左和正下）假设有m个星星，则它属于level[m]的一员，最后输出属于level[i]  （0<=i<=n-1）的星星个数。

通过做这道题，发现自己还是对线段树的含义没有太明白，稍微变变就不懂。其实树状数组最主要的两个操作就是区间更新和区间求和，题目中涉及这方面的可以想想是否能用树状数组解决。本题用的树状数组，给出的点按y排好序，相同的y按x的递增顺序给出。y不用考虑，只要考虑x就可以了。

代码中c[i] 表示横坐标为i的点的个数，由于树状数组的最小下标是从1开始的，所以在输入后，给x++，样例中 c[1]=1   c[3]=1  c[5]=2  c[7]=1

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=32002;
int level[maxn];
int c[maxn];

int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int j)
{
    while(i<=maxn)
    {
        c[i]+=j;
        i+=lowbit(i);
    }
}
int sum(int i)//1到i之间点的个数
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}

int main()
{
    int n;cin>>n;
    memset(c,0,sizeof(c));
    memset(level,0,sizeof(level));
    int x,y;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&y);
        x++;
        level[sum(x)]++;//先查询，因为算出的总个数不包括自己,sum(x)表示以前的点x小于等于当前点的x值的个数,比如个数为m,则level[m]+1,因为该点是level[m]中点个数中的一个，最后要求输出的是属于level[m]的点的个数
        add(x,1);//加上当前点
    }
    for(int i=0;i<n;i++)
    {
        printf("%d
",level[i]);
    }
    return 0;
}