POJ 2352 Stars（樹狀數組）

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22464		Accepted: 9787

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

 

 

 

可以用線段數做，也可以用樹狀數組。

還是感覺樹狀數組比較簡單一些。

 

/*
POJ 2352 Stars
就是求每個小星星左小角的星星的個數。坐標按照Y升序，Y相同X升序的順序給出
由於y軸已經排好序，可以按照x坐標建立一維樹狀數組
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAXN=15010;
const int MAXX=32010;
int c[MAXX];//樹狀數組的c數組
int cnt[MAXN];//統計結果
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=MAXX)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    int x,y;
    while(scanf("%d",&n)!=EOF)
    {
        memset(c,0,sizeof(c));
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            //加入x+1，是為了避免0，X是可能為0的
            int temp=sum(x+1);
            cnt[temp]++;
            add(x+1,1);

        }
        for(int i=0;i<n;i++)
         printf("%d\n",cnt[i]);
    }
    return 0;
}