POJ 2352 star （树状数组）

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本文介绍了一个天文学领域的算法问题，该问题关注如何高效地统计二维平面上各星星的分布级别。输入为一系列星星的坐标，输出则是不同级别的星星数量。文章提供了一种使用树状数组（Binary Indexed Tree, BIT）来优化查询和更新操作的方法。

Language:

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35318		Accepted: 15326

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

Sample Output

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

这一个题我发现了当x=0时，x+lowbit(x)=0;所以必须x++，否则update（0）就会死循环

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 32005

struct stud{
 int x,y;
 bool operator <(const stud b) const
 {
     if(x==b.x)
       return y<b.y;
     return x<b.x;
 }
}f[15005];

int ans[N];
int c[N];
int ma,n;

inline int lowbit(int x)
{
    return x&(-x);
}

void update(int x)
{
    while(x<=ma)
    {
        c[x]++;
        x+=lowbit(x);
    }
}
int Sum(int x)
{
    int s=0;
    while(x)
    {
        s+=c[x];
        x-=lowbit(x);
    }
   return s;
}

int main()
{
    int i,j;
   
    sf(n);
    mem(ans,0);
    ma=-1;
    for(i=0;i<n;i++)
    {
         sff(f[i].x,f[i].y);
         f[i].y++;
         ma=max(ma,f[i].y);
    }

    sort(f,f+n);
    mem(c,0);

    for(i=0;i<n;i++)
    {
        int t=Sum(f[i].y);
        ans[t]++;
        update(f[i].y);
    }
    for(i=0;i<n;i++)
        pf("%d\n",ans[i]);

    return 0;

}