1055 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

题意：

　　按照所给的要求排序。

思路：

　　最朴素的方法就是先按照年龄排序，然后再将所在年龄区间的人数再次排序。但是这样做的话有一组数据会超时。看了别人的blog后发现，问题出在每次查询的人数k < 100. 如果将所有满足要求年龄区间的人数都算进去的话，数字要远大于100，所以要先进行预处理，即先按照要求排序，再将让每一个年龄的人数最多有一百个。这样处理过之后，如果用cin和cout还是有一组数据过不了，换成scanf和printf就能够通过了。但是换过之后有一个比较容易卡到的点就是比较两个char数组的方法，这里我们使用strcmp()函数。

C 库函数 int strcmp(const char *str1, const char *str2) 把 str1 所指向的字符串和 str2 所指向的字符串进行比较。

该函数返回值如下：

• 如果返回值小于 0，则表示 str1 小于 str2。
• 如果返回值大于 0，则表示 str1 大于 str2。
• 如果返回值等于 0，则表示 str1 等于 str2。

Code：

#include <bits/stdc++.h>

using namespace std;

struct People {
    char name[10];
    int age;
    int netWorth;
};

bool cmp(People a, People b) {
    if (a.netWorth != b.netWorth)
        return a.netWorth > b.netWorth;
    else if (a.age != b.age)
        return a.age < b.age;
    else
        return (strcmp(a.name, b.name) < 0);
}

int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    vector<People> v(n);
    for (int i = 0; i < n; ++i)
        scanf("%s%d%d", v[i].name, &v[i].age, &v[i].netWorth);
    sort(v.begin(), v.end(), cmp);
    vector<People> temp;
    vector<int> nums(205, 0);
    for (int i = 0; i < n; ++i) {
        if (nums[v[i].age] < 100) {
            temp.push_back(v[i]);
            nums[v[i].age]++;
        }
    }
    int m, Amin, Amax, cnt = 1;
    while (cnt <= k) {
        scanf("%d%d%d", &m, &Amin, &Amax);
        vector<People> res;
        for (int i = 0; i < temp.size(); ++i) {
            if (temp[i].age >= Amin && temp[i].age <= Amax)
                res.push_back(temp[i]);
        }
        printf("Case #%d:
", cnt++);
        for (int i = 0; i < res.size() && i < m; ++i) {
            printf("%s %d %d
", res[i].name, res[i].age, res[i].netWorth);
        }
        if (res.empty()) printf("None
");
    }

    return 0;
}