PAT 甲级 1055 The World's Richest

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分类专栏： PAT甲级 PAT甲级/乙级机试经验文章标签： PAT甲级 The World's Richest 1055

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本文介绍了一个算法，用于从全球富豪名单中筛选出特定年龄段内的最富有人群。该算法需处理大量数据，通过预排序和高效查询策略，在限定条件下快速找到指定年龄段内最富有的M个人。

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1055 The World's Richest （25 point(s)）

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

经验总结：

这一题，思路上没有什么难度，但是实现流程要注意，否则第三个测试点会超时，这里可以直接暴力建立10^5的数组，然后输入数据，进行排序，然后根据查询直接输出即可AC，（稍微追求速度的话，就要在题目所给的K<=100下文章了）但是如果以年龄为键值，将相同年龄的人放入一个vector，然后再根据查询排序输出，就会超时，可能是因为测试样例查询次数十分多，每次都进行排序实在有些浪费时间，所以这一题所总结的经验就是，遇到有查询的题目，数据尽量在查询之前就已经固定，这样才可能尽量避免超时，就这么多啦~

AC代码

#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=100010;
struct people
{
	int age,worth;
	char name[10];
}p[maxn],q[maxn];
bool cmp(people a,people b)
{
	if(a.worth!=b.worth)
		return a.worth>b.worth;
	if(a.age!=b.age)
		return a.age<b.age;
	return strcmp(a.name,b.name)<0;
}

int main()
{
	int n,m,k,amin,amax,Age[210]={0};
	scanf("%d %d",&n,&k);
	for(int i=0;i<n;++i)
		scanf("%s %d %d",p[i].name,&p[i].age,&p[i].worth);
	sort(p,p+n,cmp);
	int realnum=0;
	for(int i=0;i<n;++i)
	{
		if(p[i].age<=200&&p[i].age>0&&Age[p[i].age]<100)
		{
			++Age[p[i].age];
			q[realnum++]=p[i];
		}
	}
	for(int i=0;i<k;++i)
	{
		scanf("%d %d %d",&m,&amin,&amax);
		printf("Case #%d:\n",i+1);
		int num=0;
		for(int j=0;j<realnum&&num<m;++j)
		{
			if(q[j].age>=amin&&q[j].age<=amax)
			{
				printf("%s %d %d\n",q[j].name,q[j].age,q[j].worth);
				++num;
			}
		}
		if(num==0)
			printf("None\n");
	}
	return 0;
}