Stars POJ - 2352

题目链接:点我

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    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.

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题意:

给你一群星星的坐标,要你求每个星星的左下共有多少颗星星.输出level由0 到n-1的星星的个数.

思路:

树状数组,由于题目中所有的星星是按照y的升序来输入的,所以我们只需要查询坐标小于等于它的星星共有多少颗即可.

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;

const int maxn = 32005;
int a[maxn];
int level[maxn];
int n;

void add(int x){
    while(x < maxn){
        a[x] ++;
        x += x&-x;
    }
}

int sum(int x){
    int ans = 0;
    while(x > 0){
        ans += a[x];
        x -= x&-x;
    }
    return ans;
}

int main(){
    while(scanf("%d", &n) != EOF){
        memset(a, 0, sizeof(a));
        memset(level, 0, sizeof(level));
       for(int i = 0;i < n; ++i){
        int x, y;
        scanf("%d %d", &x, &y);
        ++ x;
        level[sum(x)] ++;
        add(x);
       }
       for(int i = 0; i < n; ++ i)
        printf("%d\n",level[i]);
    }
    return 0;
}