Stars

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7209Accepted Submission(s): 2830

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
Source Ural Collegiate Programming Contest 1999
題意：給n個星星，每個星星按y坐標從小到大，y一樣x從小到大輸入，然後每個星星的做下區域每包含一個星星（不包括自己），該星星就升一級；最後求等級0~n-1的星星的個數。 分析：樹狀數組基礎入門題，也不知道怎麼說，就屬於模板題吧。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include<map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 32010

int n,a[MAXN],sum[MAXN];

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

int Sum(int n)///前n項的和
{
    int s = 0;
    while(n > 0)
    {
        s += a[n];
        n -= lowbit(n);
    }
    return s;
}

void add(int x)///增加某個元素的大小
{
    while(x <= MAXN)
    {
        ++a[x];
        x += lowbit(x);
    }
}

int main()
{
    int x,y;
    while(~scanf("%d",&n))
    {
        CL(a, 0);
        CL(sum, 0);
        for(int i=0; i<n; int="" i="0;" return="" span=""></n;></algorithm></cmath></vector></set></map></queue></stack></cstring></cstdio></iostream></span>