PAT 1055 The World's Richest

#include <cstdio>

#include <cstdlib>

#include <cstring>



#include <vector>

#include <queue>

#include <algorithm>



using namespace std;



#define AGE_MAX 200



class People {

public:

    char name[9];

    int worth;

    int age;

    int idx;

    People(const char* _name, int _worth = 0, int _age = 0) {

        strcpy(name, _name);

        worth     = _worth;

        age     = _age;

        idx     = 0; 

    }

};



bool people_compare(const People* a, const People* b) {

    if (a->worth > b->worth) {

        return true;

    } else if (a->worth < b->worth) {

        return false;

    }

    if (a->age < b->age) {

        return true;

    } else if (a->age > b->age) {

        return false;

    }

    

    return strcmp(a->name, b->name) < 0;

}



class mycmp {

public:

    bool operator() (const People* a, const People* b) {

        return !people_compare(a, b);

    }

};







int main() {

    int N = 0, K = 0;

    scanf("%d%d", &N, &K);

    

    vector<vector<People*> > peoples(AGE_MAX + 1);

    

    char name[10] = {'\0'};

    int worth = 0, age = 0;

    

    for (int i=0; i<N; i++) {

        scanf("%s%d%d", name, &age, &worth);

        peoples[age].push_back(new People(name, worth, age));

    }

    

    for (int i=0; i<=AGE_MAX; i++) {

        vector<People*>& list = peoples[i];

        if (!list.size()) continue;

        // sort people in each age list

        sort(list.begin(), list.end(), people_compare);

        

        for (int j=0; j<list.size(); j++) {

            list[j]->idx = j;

        }

    }

    

    for (int i=0; i<K; i++) {

        int M = 0, Amin = 0, Amax = 0;

        scanf("%d%d%d", &M, &Amin, &Amax);

        

        priority_queue<People*, vector<People*>, mycmp> age_leader;

        

        for(int j = Amin; j <= Amax; j++) {

            if (peoples[j].empty()) continue;

            age_leader.push(peoples[j].front());

        }

        printf("Case #%d:\n", i + 1);

        int m = 0;

        while (!age_leader.empty() && m < M) {

            m++;

            People* leader = age_leader.top();

            age_leader.pop();

            

            printf("%s %d %d\n", leader->name, leader->age, leader->worth);

            if (leader->idx + 1 >= peoples[leader->age].size()) continue;

            age_leader.push(peoples[leader->age][leader->idx + 1]);

        }

        if (m == 0) {

            printf("None\n");

        }

    }

    

    return 0;

}

室友说直接排序会超时，于是尝试着改进一下，实质上就是对有序多链表的Merge操作，这里有序链表就是以年龄划分的人群以worth等字段的排序结果，由于题目中指定最多显示的数目，这样可以不用把整个Merge做完，结果数量达到即可。一次过！