PAT A1055 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
以下是AC的代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100100;
const int len = 10;
struct person{
	char name[len];
	int age,netw;
}p[maxn];
bool cmp(person a,person b){
	if(a.netw != b.netw) return a.netw>b.netw;
	else if(a.age != b.age) return a.age<b.age;
	else return strcmp(a.name,b.name)<0;
}
int main(void){
	int n,k;
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%s%d%d",&p[i].name,&p[i].age,&p[i].netw);
	}
	sort(p,p+n,cmp);
	for(int i=0;i<k;i++){
		int total,age1,age2;
		bool fnd=false;
		scanf("%d%d%d",&total,&age1,&age2);
		printf("Case #%d:\n",i+1);
		for(int j=0;j<n;j++){
			if(p[j].age<=age2&&p[j].age>=age1&&total>0){
				fnd=true;
				printf("%s %d %d\n",p[j].name,p[j].age,p[j].netw);
				total--;
			}
		}
		if(!fnd) printf("None\n");
	}
	return 0;
}

思路：
这道题我最开始的想法是，把所有人按照年龄排序，然后按照每个query要求的年龄段再按照题目要求资产净值排序，然后从上到下输出即可。写完发现一个问题，就是做完每个query之后，因为数组不再是年龄有序的，所以必须要重新排，让数组再次年龄有序。
这样产生了一个问题，就是复杂度过高。n最大可能 10 ^ 5，每次做query 都可能要重新排，query 的 k最大 10 ^ 3，最坏情况下可能达到10 ^ 8。结果果然运行超时了，第二个和第三个数据点。接下来改进，每次查询完之后，不再还原整个数组，只还原打乱的地方。但是第三个数据点还是会超时。（这里其实最大时间复杂度没变，还是10 ^ 8)
再重新思考。如果我每次都不还原呢，只有这样时间复杂度可能通过。于是就写了上面的程序。按照题目要求排序，优先级是 资产净值>年龄>名字字典序.每次查询的时候，从上到下，遇到符合要求的输出，同时total–；这样来控制。

标签：10,PAT,A1055,int,age,query,netw,World,5888
来源： https://blog.csdn.net/weixin_48165493/article/details/113827614