poj2352-Stars

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

 

题目大意：

在坐标上有n个星星，如果某个星星坐标为(x, y), 它的左下位置为：(x0,y0)，x0<=x 且y0<=y。如果左下位置有a个星星，就表示这个星星属于level x

按照y递增，如果y相同则x递增的顺序给出n个星星，求出所有level水平的数量。

分析与总结：

因为输入是按照按照y递增，如果y相同则x递增的顺序给出的， 所以，对于第i颗星星，它的level就是之前出现过的星星中，横坐标x小于等于i星横坐标的那些星星的总数量（前面的y一定比后面的y小）。

所以，需要找到一种数据结构来记录所有星星的x值，方便的求出所有值为0～x的星星总数量。

树状数组和线段树都是很适合处理这种问题。

 

此处用树状数组来解决，要点是每次输入一颗星星，得事先算出星星的level，再将星星加入区间，以便于后续输入的星星能将此星星给算进去。

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

const int MAXN = 32005;

int c[MAXN],level[MAXN],n;
//数组c[i]=cnt表示在位置i有cnt颗星星,初始时没有星星,level[i]=cnt表示星星等级为i的星星有cnt颗。
int lowbit(int x){return x & (-x);}

int sum(int n){//计算当前星星的等级，即统计当前星星左方共有多少星星
    int sum = 0;
    for(int i=n;i>0;i-=lowbit(i))
    {
        sum += c[i];
    }
    return sum;
}
// 位置x增加了一颗星星，向上进行更新增加星星的个数
void add(int x){
    for(int i=x;i<=MAXN;i+=lowbit(i))
    {
        ++c[i];
    }
}

int main(){
    int n,x,y;
    while(~scanf("%d",&n)){
        memset(level, 0, sizeof(level));
        memset(c, 0, sizeof(c));
        for(int i=0; i<n; ++i) {
            scanf("%d%d",&x,&y);
            ++x;
            level[sum(x)]++;//sum(x)计算出星星的level,然后该level的星星个数+1
            add(x);//将这颗星星加入区间
        }
        for(int i=0; i<n; ++i)
            printf("%d\n",level[i]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/LJHAHA/p/11172590.html