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PAT 1055 The World's Richest

#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;#define AGE_MAX 200class 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