PAT (Advanced) 1055. The World's Richest (25)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

struct People
{
	string name;
	int age;
	int net_worth;
};

vector<People> people;
bool cmp(const People &a, const People &b)
{
	if (a.net_worth != b.net_worth)
		return a.net_worth > b.net_worth;
	else
	{
		if (a.age != b.age)
			return a.age < b.age;
		else
			return a.name < b.name;
	}
}
int main()
{
	int n, k;
	cin >> n >> k;
	people.resize(n);
	for (int i = 0; i < n; i++)
	{
		cin >> people[i].name;
		scanf("%d %d", &people[i].age, &people[i].net_worth);
	}
	sort(people.begin(), people.end(), cmp);
	int counts[201] = { 0 };
	int remain[100005];
	int remain_cnt = 0;
	for (int i = 0; i < n; i++)
	{
		if (++counts[people[i].age] < 101)
			remain[remain_cnt++] = i;
	}
	int m, small, big;
	for (int i = 0; i < k; i++)
	{
		int cnt = 0;
		cin >> m >> small >> big;
		cout << "Case #" << i + 1 << ":" << endl;
		int age;
		for (int j = 0; j < remain_cnt && cnt < m; j++)
		{
			age = people[remain[j]].age;
			if (age >= small && age <= big)
			{
				cnt++;
				printf("%s %d %d\n", people[remain[j]].name.c_str(), people[remain[j]].age, people[remain[j]].net_worth);
			}
		}
		if (cnt == 0)
			cout << "None" << endl;
	}
	return 0;
}

这个题目思路很简单，但是会有超时的问题。解决方法是在排序好之后，遍历一遍，如果某个年龄出现的次数超过100之后，下次碰到这个年龄就跳过。

参考：http://www.2cto.com/kf/201308/239740.html