PAT 甲级1012. The Best Rank (25)

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这道题的坑点在于需要注意输出的并列排名，比如90、60、60、50的对应排名应该是1、2、2、4，很容易搞成1、2、3、4千万注意这一点。

#include <iostream>#include <algorithm>#include <map>#include <vector>#include <string>using namespace std;struct student{    string StudentID;    int score[4];    int bestRank;    int currRank;    char bestRankLable;     student();};student::student(){    bestRank=2001;}char lables[]={'A','C','M','E'};int cmp0(const student A,const student B){    return A.score[0]>B.score[0];}int cmp1(const student A,const student B){    return A.score[1]>B.score[1];}int cmp2(const student A,const student B){    return A.score[2]>B.score[2];}int cmp3(const student A,const student B){    return A.score[3]>B.score[3];}vector <student> getBestRank(vector <student> &vS,int N,int lable){    vS[0].currRank=1;    if(vS[0].bestRank>1){        vS[0].bestRank=1;               vS[0].bestRankLable=lables[lable];    }    for(int i=1;i<N;i++){        if(vS[i].score[lable]==vS[i-1].score[lable]){            vS[i].currRank=vS[i-1].currRank;        }else{            vS[i].currRank=i+1;        }        if(vS[i].bestRank>vS[i].currRank){            vS[i].bestRank=vS[i].currRank;            vS[i].bestRankLable=lables[lable];        }    }    return vS;}int main(int argc, char** argv) {    int N,M;    map<string,student> mStudent;     vector<student> vStudent;    student temp;    string id;    cin>>N>>M;    for(int i=0;i<N;i++){        cin>>temp.StudentID>>temp.score[1]>>temp.score[2]>>temp.score[3];        temp.score[0]=(temp.score[1]+temp.score[2]+temp.score[3]);        vStudent.push_back(temp);    }    sort(vStudent.begin(),vStudent.end(),cmp0);    getBestRank(vStudent,N,0);    sort(vStudent.begin(),vStudent.end(),cmp1);    getBestRank(vStudent,N,1);    sort(vStudent.begin(),vStudent.end(),cmp2);    getBestRank(vStudent,N,2);    sort(vStudent.begin(),vStudent.end(),cmp3);    getBestRank(vStudent,N,3);    for(int i=0;i<N;i++){        mStudent[vStudent[i].StudentID]=vStudent[i];    }    for(int i=0;i<M;i++){        cin>>id;        if(mStudent.count(id)>0){            cout<<mStudent[id].bestRank<<" "<<mStudent[id].bestRankLable<<endl;        }else{            cout<<"N/A"<<endl;        }       }}

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