1055. The World's Richest (25)

最新推荐文章于 2021-08-22 16:32:31 发布

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最新推荐文章于 2021-08-22 16:32:31 发布

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分类专栏：浙大PAT 文章标签： pat vector

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本文介绍了一种算法，用于从给定的人员名单中找出特定年龄段内的最富有的人群。输入包括人员姓名、年龄和财富值，算法需处理查询请求并按财富值降序输出结果。

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时间限制

400 ms

内存限制

128000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world's wealthiest people. Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages. That is, given the net worths of N people, you must find the M richest people in a given range of their ages.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=10⁵) - the total number of people, and K (<=10³) - the number of queries. Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200]), and the net worth (integer in [-10⁶, 10⁶]) of a person. Finally there are K lines of queries, each contains three positive integers: M (<= 100) - the maximum number of outputs, and [Amin, Amax] which are the range of ages. All the numbers in a line are separated by a space.

Output Specification:

For each query, first print in a line "Case #X:" where X is the query number starting from 1. Then output the M richest people with their ages in the range [Amin, Amax]. Each person's information occupies a line, in the format

Name Age Net_Worth

The outputs must be in non-increasing order of the net worths. In case there are equal worths, it must be in non-decreasing order of the ages. If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names. It is guaranteed that there is no two persons share all the same of the three pieces of information. In case no one is found, output "None".Sample Input:

12 4
Zoe_Bill 35 2333
Bob_Volk 24 5888
Anny_Cin 95 999999
Williams 30 -22
Cindy 76 76000
Alice 18 88888
Joe_Mike 32 3222
Michael 5 300000
Rosemary 40 5888
Dobby 24 5888
Billy 24 5888
Nobody 5 0
4 15 45
4 30 35
4 5 95
1 45 50

Sample Output:

Case #1:
Alice 18 88888
Billy 24 5888
Bob_Volk 24 5888
Dobby 24 5888
Case #2:
Joe_Mike 32 3222
Zoe_Bill 35 2333
Williams 30 -22
Case #3:
Anny_Cin 95 999999
Michael 5 300000
Alice 18 88888
Cindy 76 76000
Case #4:
None

这题比较郁闷。25分钟搞定后，超时扣了4分。然后由写了半小时，换了种算法，错误还增加了。最后上网，发现别人的算法和我的基本一致，郁闷，他说有可能是因为cout用时比printf久。。

通过这题，更加深入了解了结构体的方法，还是有收获的。

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
struct node
{
public:
	string name;
	int age;
	int povety;
	node(){};
	node(string n,int a,int p)
	{
		name=n;
		age=a;
		povety=p;
	}
	void out()
	{
		cout<<name<<" "<<age<<" "<<povety<<endl;
	}
};

bool compare(node a,node b);
struct out
{
	vector<node> people;
};
int main()
{
	int K,N;
	cin>>N>>K;
	vector <node> people;
	for(int i=0;i<N;i++)
	{
		string s;
		int a,p;
		cin>>s>>a>>p;
		people.push_back(node(s,a,p));
	}
	sort(people.begin(),people.end(),compare);
	for(int i=0;i<K;i++)
	{
		cout<<"Case #"<<i+1<<":"<<endl;
		int num,low,high;
		cin>>num>>low>>high;
		int k=0;
		for(int j=0;j<N;j++)
		{
			if(k>=num) break;
			if(people[j].age<=high&&people[j].age>=low)
			{
				people[j].out();
				k++;
			}
		}
		if(k==0) cout<<"None"<<endl;
	}

	system("pause");
	return 0;	
}
bool compare(node a,node b)
{
	if(a.povety!=b.povety)
		return a.povety>b.povety;
	else if(a.age!=b.age)
		return a.age<b.age;
	return a.name<b.name;
}