题目描述

题目链接：
https://www.acwing.com/problem/content/431/

思路分析

1. 第一种方法：重载小于号
2. 第二种方法：自定义比较函数

代码实现

重载小于号

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 310;

int n;
struct Person
{
    int id, sum, a, b, c;
    bool operator< (const Person& t) const
    {
        if (sum != t.sum) return sum > t.sum;
        if (a != t.a) return a > t.a;
        return id < t.id;
    }
}q[N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ )
    {
        int a, b, c;
        cin >> a >> b >> c;
        q[i] = {i, a + b + c, a, b, c};
    }

    sort(q + 1, q + n + 1);

    for (int i = 1; i <= 5; i ++ )
        cout << q[i].id << ' ' << q[i].sum << endl;

    return 0;
}

自定义比较函数

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 310;

int n;
struct Person
{
    int id, sum, a, b, c;
}q[N];

bool cmp(Person& a, Person& b)
{
    if (a.sum != b.sum) 
    return a.sum > b.sum;
    
    if (a.a != b.a) 
    return a.a > b.a;
    
    return a.id < b.id;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ )
    {
        int a, b, c;
        cin >> a >> b >> c;
        q[i] = {i, a + b + c, a, b, c};
    }

    sort(q + 1, q + n + 1, cmp);

    for (int i = 1; i < = 5; i ++ )
        cout << q[i].id << ' ' << q[i].sum << endl;

    return 0;
}