PAT 1080 Graduate Admission[排序][难]

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 G​E​​, and the interview grade G​I​​. The final grade of an applicant is (G​E​​+G​I​​)/2. 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 G​E​​. 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 2+K integers separated by a space. The first 2 integers are the applicant's G​E​​ and G​I​​, respectively. The next K integers represent the preferred schools. For the sake of simplicity, we assume that the schools are numbered from 0 to M−1, and the applicants are numbered from 0 to N−1.

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:

0 10
3
5 6 7
2 8

1 4

题目大意：模拟考研录取情况，首先给出N个学生，学生有国家分数和面试分数，首先根据总分排名，再根据国家分数排名，如果两者均相同，那么名次就相同；共有M个学校，给出了每个学校可以招多少人；每个学生可以报K个，根据第一志愿往后选，如果第一志愿不被录取，那么就顺次第二志愿；如果有学生的排名相等，那么即使学校名额会超出预算，那么也都要把这些学生都要了！每个院校都是按排名从高到低依次录取，如果有学生K个志愿均未被录取，那么就考研失败。

//猛一看真的很复杂，需要写很长时间慢慢琢磨吧。

我的代码在pat上AC了，在牛客网上通过率为90%,

#include <iostream>
#include <vector>
#include<cstdio>
#include<algorithm>
using namespace std;
struct Stu{
    int id,ge,gi,fin;
    int rk,addmis;
    vector<int> app;
}stu[];
bool cmp(Stu &a ,Stu& b){//需要引用传参，不然会超时的。
    if(a.fin>b.fin) return true;
    else if(a.fin==b.fin&&a.ge>b.ge)return true;
    return false;
}
bool cmp2(int &a,int &b){
    return a<b;
}
int school[];
vector<int> sv[];
int main() {
    int n,m,k;
    cin>>n>>m>>k;
    for(int i=;i<m;i++){
        cin>>school[i];
    }
    int temp,avg;
    for(int i=;i<n;i++){
        cin>>stu[i].ge>>stu[i].gi;
       // avg=(stu[i].ge+stu[i].gi)/2;
        stu[i].fin=stu[i].ge+stu[i].gi;
        stu[i].id=i;
        stu[i].addmis=-;
        for(int j=;j<k;j++){
            cin>>temp;
            stu[i].app.push_back(temp);
        }
    }
    sort(stu,stu+n,cmp);
    int rk=;
    for(int i=;i<n;i++){
        stu[i].rk=rk++;
        if(i!=){//除了第一名之外都有。
            if(stu[i].fin==stu[i-].fin&&stu[i].ge==stu[i-].ge)
                stu[i].rk=stu[i-].rk;
        }
    }
//    for(int i=0;i<n;i++){
//        cout<<stu[i].id<<" "<<stu[i].fin<<" "<<stu[i].ge<<" "<<stu[i].rk<<'\n';
//    }
    for(int i=;i<n;i++){
        for(int j=;j<k;j++){//在学校中进行分配。
            if(stu[i].addmis==-&&school[stu[i].app[j]]>){//但是那个排名相同的都录取怎么做啊！
                int v=stu[i].app[j];
                sv[v].push_back(stu[i].id);
                school[v]--;
                stu[i].addmis=;//已经被录取。
                //目前还有一个问题，就是如何判断这两个学生对应地位志愿是一样的呢？
                //怎么判定当前学生已经有学校录取了呢?
                bool flag=false;
                while(stu[i+].rk==stu[i].rk&&i<n&&stu[i+].addmis==-&&stu[i+].app[j]==v){//如果当前分配到相同的学校。
                    sv[v].push_back(stu[i+].id);
                    stu[i+].addmis=;
                    if(school[v]>)school[v]--;//...你这里写成了i--,气死了。
                    i++;
                    flag=true;//又接着安排了其他学生。
                }
               if(flag)i--;break;
            }
        }
    }
    for(int i=;i<m;i++){
        sort(sv[i].begin(),sv[i].end(),cmp2);
        for(int j=;j<sv[i].size();j++){
            cout<<sv[i][j];
            if(j!=sv[i].size()-)cout<<" ";
        }
        cout<<'\n';
    }

    return ;
}

//思想就是，先对学生按要求排序，并且有一个rank，并且根据是否跟前一个人的fin和GE分数相等，排名是否和前一位同学rank相同，那么再对学生进行遍历，根据要求分配学校，并且有一个标志是addmis判断学生是否已经被安排过了。总之是比较麻烦的。。。对了，还应该给学生一个id，因为如果不给id的话，排序后会乱。。居然一开始没给安排id....

柳神代码：https://www.liuchuo.net/archives/2453

#include <iostream>
#include <vector>
#include <algorithm>
#include<cstdio>
using namespace std;
struct peo{
    int id, ge, gi, fin;
    vector<int> choice;
};
bool cmp(peo& a, peo& b) {
    if (a.fin != b.fin) return a.fin > b.fin;
    return a.ge > b.ge;
}
bool cmp2(peo& a, peo& b) {
  return a.id < b.id;
}
int main(){
    int n, m, k, quota[], cnt[] = {};
    scanf("%d%d%d", &n, &m, &k);
    vector<peo> stu(n), sch[];
    for(int i = ; i < m; i++)
        scanf("%d",&quota[i]);
    for(int i = ; i < n; i++) {
        scanf("%d%d", &stu[i].ge, &stu[i].gi);
        stu[i].id = i;//原来id也要存的对啊！！
        stu[i].fin = stu[i].ge + stu[i].gi;//使用两者的和，而没有求均值，减少误差吧。
        stu[i].choice.resize(k);
        for(int j = ; j < k; j++)
            scanf("%d", &stu[i].choice[j]);
    }
    sort(stu.begin(), stu.end(), cmp);
    for(int i = ; i < n; i++) {
        for(int j = ; j < k; j++) {
            int schid = stu[i].choice[j];//选择学校的序号。
            int lastindex = cnt[schid] - ;
           // printf("%d\n",lastindex);
            //如果还没招满，或者已经招满了（前半部分判定为false，但是和最后一名的总分数和GE分数都相等，那么也放进去。）
            if(cnt[schid] < quota[schid] || (stu[i].fin == sch[schid][lastindex].fin) && stu[i].ge == sch[schid][lastindex].ge) {
                sch[schid].push_back(stu[i]);//存放已经招收了的学生对象。
                cnt[schid]++;
                break;//如果安排了，那么就break掉内层循环，开始安排下一个学生。
            }
        }
    }
    for(int i = ; i < m; i++) {
        sort(sch[i].begin(), sch[i].end(), cmp2);
        for(int j = ; j < cnt[i]; j++) {
            if(j != ) printf(" ");
            printf("%d", sch[i][j].id);
        }
        printf("\n");
    }
    return ;
}

//安排好的学校里存的是学生的对象，这个好~

其中会有，当当前学校没有人的时候，向量下标可能会寻址到-1，

但是！如果当前学校没有人的话，那么肯定if的前半部分条件肯定是真，就不会去判断第二个条件了！！！，就不会寻址-1了！！