（PAT)1055 The World's Richest(难题,比较难的排序)

迷南。 2022-03-28 06:46 112阅读 0赞

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

解题思路:

一开始我尝试创建多个数组来储存不同年龄段的人,不过查找量比较大,两个测试用例超时了。

注意到M的范围在100以内,因此可以进行预处理,即将每个年龄中财富在前100的人筛选出放在数组中即可

(即在原来排过序的数组中删去100名以外的)

这样最多最多有10000个人在数组中,查询时,挨个遍历即可

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <string.h>
    #include <vector>
    #include <string>
    using namespace std;
    struct Person
    {
    	int age, wealth;
    	char name[20];
    }Persons[10010],validPersons[10010];
    bool cmp(Person a, Person b) {
    	if (a.wealth!= b.wealth) {
    		return a.wealth > b.wealth;
    	}
    	else {
    		if (a.age != b.age) {
    			return a.age < b.age;
    		}
    		else {
    			return strcmp(a.name, b.name) < 0;
    		}
    	}
    }
    int Ages[10010] = { 0 };  //年龄无法超过100
    int main() {
    	int N, M;
    	scanf("%d %d", &N, &M);
    	for (int i = 0; i < N; ++i) {
    		scanf("%s %d %d", &Persons[i].name, &Persons[i].age, &Persons[i].wealth);
    	}
    	sort(Persons, Persons + N, cmp);
    	int validNumber = 0;
    	for (int i = 0; i < N; ++i) {
    		if (Ages[Persons[i].age] < 100) {   //各年龄段只要筛选前100名就行了 这样100*100最多不会超过10000
    			Ages[Persons[i].age]++;
    			validPersons[validNumber++] = Persons[i];
    		}
    	}
    	for (int i = 0; i < M; ++i) {
    		int ranks, low, high;
    		scanf("%d %d %d", &ranks, &low, &high);
    		int printNumber = 0;
    		printf("Case #%d:\n", i + 1);
    		for (int j = 0; j < validNumber && printNumber < ranks; j++) {   //不能超过数组中总人数,也不能超过需要打印的人数
    			if (validPersons[j].age <= high && validPersons[j].age >= low) {
    				printf("%s %d %d\n", validPersons[j].name, validPersons[j].age, validPersons[j].wealth);
    				printNumber++;
    			}
    		}
    		if (printNumber == 0) {
    			printf("None\n");
    		}
    	}
    	system("PAUSE");
    	return 0;
    }