1055. The World's Richest (25)

题目如下：

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

题目要求按照财富>年龄>名字的顺序排列财富榜，然后按照要求的查询条件输出特定部分。

使用一个结构体，按照题目要求重载<运算符，以便排序。

名字使用char*而不是string是为了提高运行速度，有些题目会卡时间，因此养成使用char*的习惯很重要。

最后，根据查询条件，遍历排序后的vector输出满足条件的部分。

#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <string.h>

using namespace std;

struct Person{
    char *name;
    int age;
    int worth;

    bool operator < (const Person p) const{
        if(worth > p.worth) return true;
        else if(worth < p.worth) return false;
        else{ // 财富相等，比较年龄，年龄小的优先
            if(age < p.age) return true;
            else if(age > p.age) return false;
            else{ // 年龄相等，比较名字，按照字典顺序
                for(int i = 0; i < 9; i++){
                    if(name[i] < p.name[i]) return true;
                    else if(name[i] > p.name[i]) return false;
                }
                return true;
            }
        }

    }

};

int main()
{
    vector<Person> persons;
    int N,K;
    cin >> N >> K;
    persons.resize(N);
    for(int i = 0; i < N; i++){
        char *name = new char[9];
        scanf("%s%d%d",name,&(persons[i].age),&(persons[i].worth));
        persons[i].name = name;
    }
    sort(persons.begin(),persons.end());
    int maxCnt,age_begin,age_end;
    int outputCnt = 0;
    for(int i = 0; i < K; i++){
        scanf("%d%d%d",&maxCnt,&age_begin,&age_end);
        outputCnt = 0;
        printf("Case #%d:\n",i+1);
        for(int i = 0; i < N; i++){
            if(persons[i].age >= age_begin && persons[i].age <= age_end){
                printf("%s %d %d\n",persons[i].name, persons[i].age, persons[i].worth);
                outputCnt++;
                if(outputCnt >= maxCnt) break;
            }
        }
        if(outputCnt == 0) printf("None\n");
    }
    return 0;
}