acm寒假训练1月26号第八题

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本文介绍了一个天文学领域的程序设计问题，旨在通过分析恒星地图上的星体分布，计算不同等级恒星的数量。程序使用并查集算法处理星体坐标数据，实现对星体水平的高效计算。

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Time limit1000 msMemory limit32768 kBOSWindowsSource
Ural Collegiate Programming Contest 1999

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

问题链接：https://vjudge.net/contest/279644#problem/H

问题描述：输入恒星的坐标，输出各等级的恒星有多少个。

问题分析：

程序分析：

c++程序如下：

#include<cstdio>
#include<iostream>
using namespace std;
int s[1005];
int n,m;
void init()
{
    for(int i=0;i<1005;i++)
        s[i]=i;
}
int find(int x)
{
    if(s[x]!=x)
        s[x]=find(s[x]);
    return s[x];
}
void unionn(int x,int y)
{
    int a=find(x),b=find(y);
    if(a!=b)
         s[b]=a;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b,cnt=0;
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            unionn(a,b);
        }
        for(int i=1;i<=n;i++)
        {

            if(find(i)==i)
                cnt++;
        }
        printf("%d\n",cnt);
    }
    return 0;
}