1055 The World's Richest （25 分)

2019-03-28

1055 The World’s Richest （25 分)

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

1	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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 500 ms

内存限制: 128 MB

代码长度限制: 16 KB

题目大意

给出一组富翁，再给出m,amin,amax,求年龄在amin-amax之间的前m个富翁

分析

直接排序，然后遍历筛选会超时。考虑到只需要取前100个人即可，那么每个年龄段也只需要存前100个。先把所有人存储，再排序。遍历所有人，用数组book记录各年龄段的人数，超过100的不要。用一个新的vector存放筛选过的人，最后只需到筛选过的vector中遍历即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;
struct richer{
    string name;
    int age;
    int worth;

    richer(const string &name, int age, int worth) : name(name), age(age), worth(worth) {}
};
bool cmp(richer r1,richer r2){
    return r1.worth!=r2.worth?r1.worth>r2.worth:(r1.age!=r2.age?r1.age<r2.age:r1.name<r2.name);
}
int book[201];
int main() {
    int n,k;
    scanf("%d %d",&n,&k);
    string name;
    int age,worth;
    vector<richer> billionaires;
    for(int i=0;i<n;i++) {
        cin>>name;
        scanf("%d %d",&age,&worth);
        richer r(name,age,worth);
        billionaires.push_back(r);
    }
    sort(billionaires.begin(),billionaires.end(),cmp);

    vector<richer> candidates;
    for(int i=0;i<n;i++){
        if(book[billionaires[i].age]<=100){
            book[billionaires[i].age]++;
            candidates.push_back(billionaires[i]);
        }
    }
    int m,lage,uage;
    for(int i=1;i<=k;i++){


        printf("Case #%d:\n",i);
        scanf("%d %d %d",&m,&lage,&uage);
        int t=m;
        for(auto r:candidates){
            if(r.age>=lage&&r.age<=uage){
                printf("%s %d %d\n",r.name.c_str(),r.age,r.worth);
                m--;
            }
            if(m==0)
                break;
        }
        if(t==m)
            printf("None\n");
    }
    return 0;
}