1055 The World‘s Richest

Name Age Net_Worth

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

题目大意

给你N个人的信息 : name age rich

查询K次

每次输出年龄段[left，right]内，rich排前X名的人，rich相同比较age，age相同比较name

思路

见代码

C/C++ 

 
#include
using namespace std;
struct People{
    string name;
    int age,rich;
}p;
vector age[201],result;
bool cmp(const People& x,const People& y){
    return x.rich != y.rich ? x.rich > y.rich : x.age!=y.age ? x.age
}
int main()
{
 
 
    int N,M,left,right;
    cin >> N >> M;
    while (N--){
        cin >> p.name >> p.age >> p.rich;
        age[p.age].push_back(p);
    }
 
    for(int z=1;z<=200;z++) {
        sort(age[z].begin(),age[z].end(),cmp);
        while (age[z].size()>100) age[z].pop_back();
    }
 
    for(int z=1;z<=M;z++){
        printf("Case #%d:\n",z);
        cin >> N >> left >> right;
 
        while (left<=right){
            for(const People& x:age[left]) result.push_back(x);
            left++;
            if(result.size()>=N) break;
        }
 
        sort(result.begin(),result.end(),cmp);
        while (result.size()>N) result.pop_back();
 
        while (left<=right){
            for(const People& x: age[left]){
                if(x.rich<=result.back().rich) break;
                int flag = N;
                while (flag>0 && x.rich > result[flag-1].rich) flag--;
                result.insert(result.begin()+flag,x);
                result.pop_back();
            }
            left++;
        }
 
        for(const People& x: result) cout << x.name << " " << x.age << " " << x.rich << endl;
        if(result.empty()) puts("None");
        else result.clear();
    }
    return 0;
}