PAT A1055

• 題目：

  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

• 解題思路

  自己思路：

  首先對年齡排序，再對輸入的年齡區間按照要求排序。然後再輸出這個區間前M個值。

  代碼實作：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int Max = 100010;
struct People{
    char name[10];
    int age;
    int worth;
} peo[Max];
bool cmp1(People a, People b){
    return a.age < b.age;
}
bool cmp2(People a, People b){
    if(a.worth != b.worth)
        return a.worth > b.worth;                 // 按照财富從大到小排序
    else if(a.age != b.age)
        return a.age < b.age;
    return strcmp(a.name, b.name) < 0;
}
int main()
{
    int N, K;
    cin >> N >> K;
    for(int i =0; i < N; i++){
        scanf("%s%d%d", peo[i].name, &peo[i].age, &peo[i].worth);
    }
    for(int k= 1; k <= K; k++){
        int Max_num, min_age, max_age, count1 = 0, j = Max;
        cin >> Max_num >> min_age >> max_age;
        sort(peo, peo+N, cmp1);                  //按照年齡排序
        for(int i = 0; i < N; i++){
            if(peo[i].age >= min_age && peo[i].age <=max_age){
                count1++;
                if(j > i)                   //獲得最開始位置的值
                    j = i;
            }
        }
        sort(peo+j, peo+j+count1, cmp2);        //在這個年齡區間按照規則排序
        if(count1 == 0){
            printf("Case #%d:\n", k);
            printf("None\n");
        }
        else{
            printf("Case #%d:\n", k);
            for(int a = j; a < j+ min(Max_num,count1); a++)
                printf("%s %d %d\n",peo[a].name, peo[a].age, peo[a].worth);
        }
    }
    return 0;
}

           

這樣做測試點2，3會發生逾時。這裡排序了兩次，思考有沒有可能隻用排序一次。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int Max = 100010;
struct People{
    char name[10];
    int age;
    int worth;
} peo[Max];
bool cmp(People a, People b){
    if(a.worth != b.worth)
        return a.worth > b.worth;                 // 按照财富從大到小排序
    else if(a.age != b.age)
        return a.age < b.age;
    return strcmp(a.name, b.name) < 0;
}
int main()
{
    int N, K;
    cin >> N >> K;
    for(int i =0; i < N; i++){
        scanf("%s%d%d", peo[i].name, &peo[i].age, &peo[i].worth);
    }
    sort(peo, peo+N, cmp);             //排序一次
    for(int k= 1; k <= K; k++){
        int Max_num, min_age, max_age, count1 = 0,j = Max;
        bool tag = false;
        cin >> Max_num >> min_age >> max_age;
         printf("Case #%d:\n",k);
        for(int i = 0; i < N; i++){
          if(peo[i].age >= min_age && peo[i].age <=max_age &&count1 < Max_num ){
                printf("%s %d %d\n",peo[i].name, peo[i].age, peo[i].worth);
                tag = true;
                count1++;
            }
        }
        if(!tag)
            printf("None\n");

    }
    return 0;
}

           

這次隻有第二處測試點運作逾時。發現隻要把最後一處if判斷語句裡面Max_num > 0放到for循環裡面即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int Max = 100010;
struct People{
    char name[9];
    int age;
    int worth;
} peo[Max];
bool cmp(People a, People b){
    if(a.worth != b.worth)
        return a.worth > b.worth;                 // 按照财富從大到小排序
    else if(a.age != b.age)
        return a.age < b.age;
    return strcmp(a.name, b.name) < 0;
}
int main()
{
    int N, K;
    scanf("%d%d",&N,&K);
    for(int i =0; i < N; i++){
        scanf("%s%d%d", peo[i].name, &peo[i].age, &peo[i].worth);
    }
    sort(peo, peo+N, cmp);
    for(int k= 1; k <= K; k++){
        int Max_num, min_age, max_age;
        bool tag = false;
        scanf("%d%d%d",&Max_num, &min_age, &max_age);
         printf("Case #%d:\n",k);
        for(int i = 0; i < N &Max_num > 0; i++){
            if(peo[i].age >= min_age && peo[i].age <=max_age ){
                printf("%s %d %d\n",peo[i].name, peo[i].age, peo[i].worth);
                tag = true;
               Max_num--;
            }
        }
        if(!tag)
            printf("None\n");

    }
    return 0;
}