PAT甲级——A1055 The World's Richest

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 (≤) - the total number of people, and K (≤) - 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 [−]) of a person. Finally there are K lines of queries, each contains three positive integers: M (≤) - 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

第一个代码是用vector存储，但遍历超时，第二个时使用数组存储，测试通过.
搞不明白同样的算法复杂度，为什么vector遍历时间就比数组差那么多？

 #include <iostream>
 #include <vector>
 #include <string>
 #include <algorithm>
 using namespace std;
 struct Node
 {
     string name;
     int age, val;
 }node;
 int N, K, M, num, Amin, Amax;
 bool cmp(Node a, Node b)
 {
     if (a.val == b.val && a.age == b.age)
         return a.name < b.name;
     else if (a.val == b.val)
         return a.age < b.age;
     else
         return a.val > b.val;
 }
 int main()
 {
     cin >> N >> K;
     vector<Node>v;
     vector<vector<Node>>res(K);
     for (int i = ; i < N; ++i)
     {
         cin >> node.name >> node.age >> node.val;
         v.push_back(node);
     }
     sort(v.begin(), v.end(), cmp);
     for (int i = ; i < K; ++i)
     {
         cin >> num >> Amin >> Amax;
         int flag = ;
         cout << "Case #" << i +  << ":" << endl;
         for (auto a : v)
         {
             if (a.age >= Amin && a.age <= Amax)
             {
                 cout << a.name << " " << a.age << " " << a.val << endl;
                 flag++;
                 if (flag == num)
                     break;
             }
         }
         if (flag == )
             cout << "None" << endl;
     }
     return ;
 }

 #include<bits/stdc++.h>
 using namespace std;
 struct Person{//存储相应信息的结构体
     string name;
     int age,worth;
 };
 Person person[(int)(1e5+)];
 int main(){
     int N,K;
     scanf("%d%d",&N,&K);
     for(int i=;i<N;++i)
         cin>>person[i].name>>person[i].age>>person[i].worth;
     sort(person,person+N,[](const Person&p1,const Person&p2){//比较函数
         if(p1.worth!=p2.worth)
             return p1.worth>p2.worth;
         else if(p1.age!=p2.age)
             return p1.age<p2.age;
         else
             return p1.name<p2.name;
     });//排序
     for(int i=;i<=K;++i){
         int M,Amin,Amax;
         bool f=false;
         scanf("%d%d%d",&M,&Amin,&Amax);
         printf("Case #%d:\n",i);
         for(int j=;j<N&&M>;++j)//遍历数组
             if(person[j].age>=Amin&&person[j].age<=Amax){//找到符合要求的人
                 printf("%s %d %d\n",person[j].name.c_str(),person[j].age,person[j].worth);//输出
                 --M;//将M递减
                 f=true;//表示已输出过
             }
         if(!f)//在该年龄段没有任何输出
             printf("None\n");//输出None
     }
     return ;
 }