PAT1080 Graduate Admission (30)

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题目链接：

http://www.nowcoder.com/pat/5/problem/4310

题目描述：

（1）分数分为Ge和Gi两部分，按照总分给学生排名次，如果总分一样，则按照Ge来排；
（2）每个学生都有几个志愿，按照志愿高低给学生分配学校；
（3）对于名次一样又报了同一学校的学生，该学校只要接收了一个就得接收全部，即便超出了原定的招生名额；

题目分析：

每次模拟题写得总是很冗长的感觉啊。
需要注意的是写学校选学生那部分。

代码：

#include<iostream>#include<vector>#include<string.h>#include<algorithm>#include<set>using namespace std;struct Student{    int stuID;    int GE;    int GI;    int EI;    int rank;    vector<int> preferred;};bool operator <(const Student& stu1,const Student& stu2){    return stu1.EI == stu2.EI ? stu1.GE>stu2.GE:stu1.EI > stu2.EI;}struct School{    int quota;    int lowest;    set<int> admission;};int main(){    int N, M, K;    scanf("%d%d%d",&N,&M,&K);    Student* graduates = new Student[N];    School* schools = new School[M];    for (int i = 0; i < M; i++){        scanf("%d",&schools[i].quota);    }    for (int i = 0; i < N; i++){        scanf("%d%d",&graduates[i].GE,&graduates[i].GI);        graduates[i].EI = graduates[i].GE + graduates[i].GI;        graduates[i].stuID = i;        int prefer;        for (int j = 0; j < K;j++){            scanf("%d",&prefer);            graduates[i].preferred.push_back(prefer);        }    }    sort(graduates,graduates+N);    for (int i = 0; i < N; i++){        if (i>0 && graduates[i].EI == graduates[i - 1].EI && graduates[i].GE == graduates[i - 1].GE){            graduates[i].rank = graduates[i - 1].rank;        }        else{            graduates[i].rank = i + 1;        }        for (int j = 0; j < K; j++){            int schID = graduates[i].preferred[j];            if (schools[schID].quota>0 || graduates[i].rank == schools[schID].lowest){                schools[schID].lowest = graduates[i].rank;                schools[schID].quota--;                schools[schID].admission.insert(graduates[i].stuID);                break;            }        }    }    for (int i = 0; i < M; i++){        if (schools[i].admission.empty()){            printf("\n");        }        else{            auto it = schools[i].admission.begin();            printf("%d",*it++);            for (;it!=schools[i].admission.end(); it++){                printf(" %d",*it);            }            printf("\n");        }    }    return 0;}

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