傳送門
https://www.patest.cn/contests/pat-a-practise/1012
題目
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 Algebra), 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
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
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
分析
先理解本題題意：
本題要求的是學生的最佳排名，怎么叫最佳排名呢？例如：C排第3，M排第1，那么輸出的就是1 M，如果有并列情況，按照A > C > M > E的優先級處理。
因為studentId和排名，以及studentId和課程，滿足key-value關系，所以這里選用map來解題。若使用數組的話，太耗費空間。
開始先定義結構體數組，存儲studentId和各類課程成績以及總分（平均分也可以），然后讀入數據。
接著，按照A > C > M > E的順序為結構體數組排序，要單獨寫4個排序函數，每次排序后遍歷數組，若排名比之前科目的排名靠前，則將數組的下標作為排名放到map里面，更新排名。（這里會出現問題，后面會詳細說）
最后，按照輸入的studentId，從map里面取出最佳排名和對應課程名輸出即可，若未找到改學生，則輸出N/A。
遇到的問題：
要考慮到學生成績相同的情況，也就是并列排名時，學生的排名應為：
1 2 3 3 5，而不是：1 2 3 4 5。如果單純用數組的下標作為排名的話會出現問題，我想到的解決方法是引入兩個變量，用來計算當前排名以及當前分數。
若后面學生分數和前面學生分數一致（即后面學生分數 = 當前分數），則后面學生的排名也要和前面學生排名保持一致（即后面學生排名 = 當前學生排名）。
若不一致，則更新當前的排名 = 數組下標，更新當前的分數 = 學生分數。
源代碼
//C/C++實現
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
struct student{
int id;
int C;
int M;
int E;
int sum;
};
bool sortByA(const student& a, const student& b){
return a.sum > b.sum;
}
bool sortByC(const student& a, const student& b){
return a.C > b.C;
}
bool sortByM(const student& a, const student& b){
return a.M > b.M;
}
bool sortByE(const student& a, const student& b){
return a.E > b.E;
}
int main(){
int n, m;
scanf("%d %d", &n, &m);
vector<student> v;
for(int i = 0; i < n; ++i){
student s;
cin >> s.id >> s.C >> s.M >> s.E;
s.sum = s.C + s.M + s.E;
v.push_back(s);
}
map<int, int> mRank;
map<int, char> mCourse;
int currentRank, currentScore;
//sort by A(sum)
sort(v.begin(), v.end(), sortByA);
currentRank = -1, currentScore = -1;
for(int i = 0; i < v.size(); ++i){
if(v[i].sum != currentScore){
currentRank = i + 1;
}
mRank[v[i].id] = currentRank;
mCourse[v[i].id] = 'A';
currentScore = v[i].sum;
}
//sort by C
sort(v.begin(), v.end(), sortByC);
currentRank = -1, currentScore = -1;
for(int i = 0; i < v.size(); ++i){
if(v[i].C != currentScore){
currentRank = i + 1;
}
if(currentRank < mRank[v[i].id]){
mRank[v[i].id] = currentRank;
mCourse[v[i].id] = 'C';
}
currentScore = v[i].C;
}
//sort by M
sort(v.begin(), v.end(), sortByM);
currentRank = -1, currentScore = -1;
for(int i = 0; i < v.size(); ++i){
if(v[i].M != currentScore){
currentRank = i + 1;
}
if(currentRank < mRank[v[i].id]){
mRank[v[i].id] = currentRank;
mCourse[v[i].id] = 'M';
}
currentScore = v[i].M;
}
//sort by E
sort(v.begin(), v.end(), sortByE);
currentRank = -1, currentScore = -1;
for(int i = 0; i < v.size(); ++i){
if(v[i].E != currentScore){
currentRank = i + 1;
}
if(currentRank < mRank[v[i].id]){
mRank[v[i].id] = i + 1;
mCourse[v[i].id] = 'E';
}
currentScore = v[i].E;
}
//output
int tmp;
for(int i = 0; i < m; ++i){
scanf("%d", &tmp);
if(mRank.count(tmp)){
printf("%d %c\n", mRank[tmp], mCourse[tmp]);
}
else{
printf("N/A\n");
}
}
return 0;
}
