【PAT】 A1055 The Worlds Richest (25分)-优快云博客

本文链接：https://blog.youkuaiyun.com/jetpack_/article/details/106560833

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。算法首先按财富值、年龄和姓名字典序进行排序，然后遍历数组输出符合条件的人选。输入包括总人数、查询次数、个人详细信息及查询条件，输出则是按要求排序后的富豪名单。

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1055 The World’s Richest (25分)

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 (≤10 5 ) - the total number of people, and K (≤10 3 ) - 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 [−10 6 ,10 6 ]) 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

思路与总结：

先按要求1.财富值由大到小 2年龄有小到大 3名字字典序有小到大；写好cmp（）用sort（）排好
遍历结构体数组，把符合年龄要求的按序输出即可

AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int const maxn = 100010;
struct node
{
	char name[10];
	int age;
	int worth;
}man[maxn];

bool cmp(node A, node B){
	if(A.worth!=B.worth) return A.worth>B.worth;
	else if(A.age!=B.age) return A.age<B.age;
	else return strcmp(A.name, B.name)<0 ;
}
int n,k;

int main()
{
	scanf("%d%d",&n,&k);
	for(int i=0; i<n; i++){
		scanf("%s %d %d",&man[i].name, &man[i].age, &man[i].worth);
	}
	sort(man, man+n, cmp);
    
	for(int i=1; i<=k; i++){
		int m, Amin, Amax, cnt = 0;
        scanf("%d%d%d",&m,&Amin,&Amax);
		printf("Case #%d:\n",i );
		for(int j=0; j<n; j++){
			if(man[j].age>=Amin && man[j].age<=Amax){
				printf("%s %d %d\n",man[j].name, man[j].age, man[j].worth );
				cnt++;
			}
			if(cnt == m) break;
		}
		if(cnt == 0) printf("None\n");

	}
    return 0;
}