POJ 2352 Stars(簡單樹狀陣列)

畫船聽雨發表於2014-02-22
原文網址 : https://blog.csdn.net/fulongxu/article/details/19702745
陣列

什麼是樹狀陣列在這裡就不說了啊，這裡有解釋了啊。http://blog.csdn.net/fulongxu/article/details/19701281

就是一個模版題目，直接套模版都可以過。先建陣列，再求和。

由於y座標是升序的且座標不重複，所以在星星A後面輸入的星星的x,y座標不可能都小於等於星星A。假如當前輸入的星星為（3,3），易得我們只需要去找樹狀陣列中小於等於3的值就可以了。

Stars
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 29336   Accepted: 12829

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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 50000;

using namespace std;

int c[maxn];
int vis[maxn];

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

void add(int x)
{
    while(x < maxn)
    {
        c[x]++;
        x += lowbit(x);
    }
}

int sum(int x)
{
    int cnt = 0;
    while(x > 0)
    {
        cnt += c[x];
        x -= lowbit(x);
    }
    return cnt;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int x, y;
        memset(vis, 0 , sizeof(vis));
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d",&x, &y);
            x++;
            y++;
            vis[sum(x)]++;
            add(x);
        }
        for(int i = 0; i < n; i++)
            cout<<vis[i]<<endl;
    }
    return 0;
}


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