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！

PAT 1055 The World's Richest