PAT A1055. The World's Richest (25)_算法笔记a1055-优快云博客

题目链接

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

本题虽然简单，但是不能忽视其陷阱。

在第二审查点的时候，运行时间会提高，到达207ms，但不会超时（<=400ms）。原代码如下：

#include<cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
struct Del {
    char name[10];
    int age, worth;
};
Del del[100005];
bool cmp(Del a, Del b) {
    int s = strcmp(a.name, b.name);
    if(a.worth!=b.worth)return a.worth > b.worth;
    else if (a.age != b.age)return a.age < b.age;
    else return s < 0;
}
int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) {
        scanf("%s %d %d", del[i].name, &del[i].age, &del[i].worth);
    }
    sort(del, del + n, cmp);
    for (int i = 0; i < k; i++) {
        int m, amin, amax;
        scanf("%d %d %d", &m, &amin, &amax);
        int cnt = 0,age;
        printf("Case #%d:\n", i + 1);
        for (int j = 0; j < n; j++) {
            if (cnt == m)break;
            age = del[j].age;
            if (age >= amin&&age <= amax) {
                printf("%s %d %d\n", del[j].name, age, del[j].worth);
                cnt++;
            }
        }
        if (cnt == 0)printf("None\n");
    }
    return 0;
}

但是算法笔记上面运行时间要求在200ms以内。所以按照书上的思想，将数组进行了预处理，复杂度得到降低，结果在检查点2所得到的结果是150ms，其余各检查点也都有所降低，代码修改如下：

#include<cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
struct Del {
    char name[10];
    int age, worth;
};
int age[100005];
Del del[100005],ano[100005];
bool cmp(Del a, Del b) {
    int s = strcmp(a.name, b.name);
    if(a.worth!=b.worth)return a.worth > b.worth;
    else if (a.age != b.age)return a.age < b.age;
    else return s < 0;
}
int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) {
        scanf("%s %d %d", del[i].name, &del[i].age, &del[i].worth);
    }
    sort(del, del + n, cmp);
    int num = 0;
    for (int i = 0; i < n; i++) {
        if (age[del[i].age] <= 100) {
            age[del[i].age]++;
            ano[num++] = del[i];
        }
    }
    for (int i = 0; i < k; i++) {
        int m, amin, amax;
        scanf("%d %d %d", &m, &amin, &amax);
        int cnt = 0,age;
        printf("Case #%d:\n", i + 1);
        for (int j = 0; j < num; j++) {
            if (cnt == m)break;
            age = ano[j].age;
            if (age >= amin&&age <= amax) {
                printf("%s %d %d\n", ano[j].name, age, ano[j].worth);
                cnt++;
            }
        }
        if (cnt == 0)printf("None\n");
    }
    return 0;
}

增加了控制每个年龄个数的数组age。由于m<=100，所以，只需要将每个年龄上的前100名计入ano结构体数组中即可。此时，当n到达100000时，代码在经过检查点的时候，不会像第一个代码那样，遍历100000次，而是遍历20000次，有效降低了查询时的复杂度，同时也让我得到了宝贵的经验。