hdu 1541 Stars(排序＋树状数组)

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5401    Accepted Submission(s): 2134

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

51 15 17 13 35 5

 

Sample Output

12110

 

Source

Ural Collegiate Programming Contest 1999

题目分析：
可以先对横坐标排序，按照这个顺序添加，然后每次统计y坐标符合条件的数量，利用树状数组为何区间和
 

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#define MAX 15007#define N 32002using namespace std;int n;struct Star{    int x,y,l;    bool operator < ( const Star & a ) const     {        return x < a.x;    }}p[MAX];int c[N];int ans[MAX];int lowbit ( int x ){    return x&-x;}void add ( int x ){    while ( x < N  )    {        c[x]++;        x += lowbit(x);    }}int sum ( int x ){    int ret = 0;    while ( x )    {        ret += c[x];        x -= lowbit (x);    }    return ret;}int main ( ){       while  ( ~scanf ( "%d" , &n ) )    {        memset ( c , 0 , sizeof ( c ) );        memset ( ans , 0 , sizeof ( ans ) );        for ( int i = 0 ; i < n ; i++ )        {            scanf ( "%d %d" , &p[i].x , &p[i].y );            p[i].x++,p[i].y++;        }        sort ( p , p+n );        int id1,id2;        id1 = id2 = 0;        while ( id1 < n )        {            int x = p[id1].x;            int y = p[id1].y;            while ( id2 < n && p[id2].x <= x )            {                add ( p[id2].y );                id2++;            }            p[id1++].l = sum (y)-1;        }        for ( int i = 0 ; i < n ; i++ )            ans[p[i].l]++;        for ( int i = 0 ; i < n ; i++ )            printf ( "%d\n" , ans[i] );    }}