【解析】1012 The Best Rank (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks – that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID C M E A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91
Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input Specification:
Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (≤2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output Specification:
For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output N/A.

Sample Input:
5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output:
1 C
1 M
1 E
1 A
3 A
N/A

注意：如果两个分数相同，他们都是同一名，例如：100，100，90，为1、1、3。

#include<bits/stdc++.h>
using namespace std;
string intTochar="ACME";//将数组下标映射到相应字符
typedef pair<int,double> StudentScore;//存储学生id和分数的结构体
unordered_map<int,array<int,4>>studentRank;//存储学生id和4门成绩的排名，4门成绩排名存储在一个数组中，下标0123分别代表ACME四个成绩排名
int N,MM,id;//学生总数和要查询的学生数量
void setRank(const vector<StudentScore>&temp,int index){//找出学生在某一成绩上的排名，index值为0123，分别代表ACME四个成绩排名studentRank[temp[0].first][index]=1;for(int i=1;i<temp.size();++i){if(temp[i].second!=temp[i-1].second)studentRank[temp[i].first][index]=i+1;elsestudentRank[temp[i].first][index]=studentRank[temp[i-1].first][index];}
}
int main(){scanf("%d%d",&N,&MM);vector<StudentScore>ACME[4];//下标为0123的vector分别存储ACME四个成绩for(int i=0;i<N;++i){double A[4];//下标为0123的元素分别临时存储ACME四个成绩scanf("%d%lf%lf%lf",&id,&A[1],&A[2],&A[3]);A[0]=(A[1]+A[2]+A[3])/3;for(int i=0;i<4;++i)ACME[i].push_back({id,A[i]});}for(int i=0;i<4;++i)//进行排序sort(ACME[i].begin(),ACME[i].end(),[](const StudentScore&s1,const StudentScore&s2){//比较函数，按分数由大到小排序，分数相同按id从小到大排序return s1.second!=s2.second?s1.second>s2.second:s1.first<s2.first;});for(int i=0;i<4;++i)setRank(ACME[i],i);while(MM--){scanf("%d",&id);if(studentRank.find(id)==studentRank.cend()){//如果没有这样的学生，输出N/Aprintf("N/A\n");}else{//如果有这样的学生auto i=min_element(studentRank[id].begin(),studentRank[id].end());//标准库函数，按ACME优先级顺序找到排名最靠前的迭代器printf("%d %c\n",*i,intTochar[i-studentRank[id].begin()]);//输出结果}}return 0;
}

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【解析】1012 The Best Rank (25 分)

立志用最少的代码做最高效的表达

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