1080 Graduate Admission (30 分)

1080 Graduate Admission 30 分

It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admission procedure.

Each applicant will have to provide two grades: the national entrance exam grade , and the interview grade . The final grade of an applicant is . The admission rules are:

The applicants are ranked according to their final grades, and will be admitted one by one from the top of the rank list.
If there is a tied final grade, the applicants will be ranked according to their national entrance exam grade . If still tied, their ranks must be the same.
Each applicant may have K choices and the admission will be done according to his/her choices: if according to the rank list, it is one's turn to be admitted; and if the quota of one's most preferred shcool is not exceeded, then one will be admitted to this school, or one's other choices will be considered one by one in order. If one gets rejected by all of preferred schools, then this unfortunate applicant will be rejected.
If there is a tied rank, and if the corresponding applicants are applying to the same school, then that school must admit all the applicants with the same rank, even if its quota will be exceeded.

Input Specification:

Each input file contains one test case.

Each case starts with a line containing three positive integers: N (≤40,000), the total number of applicants; M (≤100), the total number of graduate schools; and K (≤5), the number of choices an applicant may have.

In the next line, separated by a space, there are M positive integers. The i-th integer is the quota of the i-th graduate school respectively.

Then N lines follow, each contains integers separated by a space. The first 2 integers are the applicant's and , respectively. The next integers represent the preferred schools. For the sake of simplicity, we assume that the schools are numbered from 0 to , and the applicants are numbered from to .

Output Specification:

For each test case you should output the admission results for all the graduate schools. The results of each school must occupy a line, which contains the applicants' numbers that school admits. The numbers must be in increasing order and be separated by a space. There must be no extra space at the end of each line. If no applicant is admitted by a school, you must output an empty line correspondingly.

Sample Input:

11 6 3
2 1 2 2 2 3
100 100 0 1 2
60 60 2 3 5
100 90 0 3 4
90 100 1 2 0
90 90 5 1 3
80 90 1 0 2
80 80 0 1 2
80 80 0 1 2
80 70 1 3 2
70 80 1 2 3
100 100 0 2 4

Sample Output:

思路

建立一个结构体用于存储学生的信息，建立一个结构体数组存储所有的学生。

struct student
{
    int ge;
    int gi;
    int choices[maxk]; // 记录学生的志愿。
    int key; // 记录学生的id，因为之后会进行排序，学生的索引便不再是他们的id了。
} Stu[maxn];

建立一个vector schAdm[maxm]; 记录每个学校录取的学生。
设计一个排序函数，将学生按分数从高到低排序。

bool cmp(student s1, student s2) {
    if((s1.ge + s1.gi) != (s2.ge + s2.gi)) {
        return (s1.ge + s1.gi) > (s2.ge + s2.gi);
    }
    else
        return s1.ge > s2.ge;
}

排序后，遍历结构体数组，将学生添加到对应学校中（如果没有满额），如果满额了，则与该学校录取的最后一名学生的成绩进行比较，如果成绩完全一样，则超额录取。否则，遍历该学生之后的志愿。

注意点

如何设计满额后也可以判断是否继续录取学生的功能。

// 当学校满员后调用该函数，判断时候可以超额录取。
bool isExceeded(int stuid, int schid) {
    int last_stu_id = schAdm[schid][schQuota[schid]-1]; // 获取学校的学生中最后一名的索引
    if((Stu[last_stu_id].ge == Stu[stuid].ge) && (Stu[last_stu_id].gi ==  Stu[stuid].gi))
        return true;
    else return false;

}

AC代码

#include 
#include
#include
using namespace std;
const int maxn = 40001;
const int maxm = 101;
const int maxk = 5;

struct student
{
    int ge;
    int gi;
    int choices[maxk];
    int key;
} Stu[maxn];

// 排序函数
bool cmp(student s1, student s2) {
    if((s1.ge + s1.gi) != (s2.ge + s2.gi)) {
        return (s1.ge + s1.gi) > (s2.ge + s2.gi);
    }
    else
        return s1.ge > s2.ge;
}

int schQuota[maxm]; // 每个学校接受学生的名额上限
vector schAdm[maxm]; // 记录每个学校录取的学生
// 判断是否可以超限添加（当两者的分数完全一样，学校必须接收，无论名额是否超限）
bool isExceeded(int stuid, int schid) {
    int last_stu_id = schAdm[schid][schQuota[schid]-1];
    if((Stu[last_stu_id].ge == Stu[stuid].ge) && (Stu[last_stu_id].gi ==  Stu[stuid].gi))
        return true;
    else return false;

}

int main() {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);

    for (int i = 0; i < m; ++i)
    {
        scanf("%d", &schQuota[i]);
    }

    for (int i = 0; i < n; ++i)
    {
        int ge, gi;
        scanf("%d%d", &ge, &gi);
        Stu[i].ge = ge;
        Stu[i].gi = gi;
        Stu[i].key = i;
        for(int j = 0; j < k; j++) {
            int choice;
            scanf("%d", &choice);
            Stu[i].choices[j] = choice;
        }
    }

    sort(Stu, Stu+n, cmp);


    for (int i = 0; i < n; ++i)
    {
        int* choices;
        choices = Stu[i].choices;
        for (int j = 0; j < k; ++j)
        {
            //printf("schAdm[%d]: %d %d\n", choices[j], schAdm[choices[j]].size(), schQuota[choices[j]]);
            if(schAdm[choices[j]].size() < schQuota[choices[j]]) {
                //int stuid = Stu[i].key;
                schAdm[choices[j]].push_back(i); // 记录的是排序后学生所在数组的序号，并不是学生真正的id
                break;
            }
            else {
                if(isExceeded(i, choices[j])) {
                    schAdm[choices[j]].push_back(i);
                    break;
                }

            }
        }
    }

    for (int i = 0; i < m; ++i)
    {
        vector adm = schAdm[i]; // 因为记录的是排序后在数组中的序号，这里获取原来的id
        vector ids;
        for (int j = 0; j < adm.size(); ++j)
        {
            int id = Stu[adm[j]].key;
            ids.push_back(id);
        }
        sort(ids.begin(), ids.end());
        for (int j = 0; j < ids.size(); ++j)
        {
            printf("%d", ids[j]);
            if(j != ids.size()-1) printf(" ");
        }
        printf("\n");
    }
}

1080 Graduate Admission (30 分)