PAT甲级 1012 The Best Rank

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

我用了两个结构体，不熟练的也可以用二维数组，具体注释放代码里了，一开始21分，后来看了大神的解释懂了，不同人的名次可能相同，之后的名次不能顺延，需要跳位，例如：

90 88 88 75

排名应该为：

1 2 2 4

PAT是真的考验代码的严谨性，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e3+5;struct Rank{char c;//分数对应的科目int score,r;//分数和该分数所在排名
};struct node{string id;Rank s[4];//四门科目成绩
}stu[N];bool cmp1(node a,node b){//比较‘A’科目的成绩return a.s[0].score>b.s[0].score;
}bool cmp2(node a,node b){//比较‘C’科目的成绩return a.s[1].score>b.s[1].score;
}bool cmp3(node a,node b){//比较‘M’科目的成绩return a.s[2].score>b.s[2].score;
}bool cmp4(node a,node b){//比较‘E’科目的成绩return a.s[3].score>b.s[3].score;
}bool cmp5(Rank a,Rank b){//比较每门成绩的排名if(a.r!=b.r) return a.r<b.r;else return a.c<b.c;//若排名相同则比较科目优先级
}int main(){int n,q;cin>>n>>q;for(int i=0;i<n;i++){cin>>stu[i].id>>stu[i].s[1].score>>stu[i].s[2].score>>stu[i].s[3].score;stu[i].s[0].score=(stu[i].s[1].score+stu[i].s[2].score+stu[i].s[3].score)/3;}sort(stu,stu+n,cmp1);for(int i=0;i<n;i++){if(i==0) stu[i].s[0].r=i+1;else if(i>0){if(stu[i].s[0].score==stu[i-1].s[0].score) stu[i].s[0].r=stu[i-1].s[0].r;else stu[i].s[0].r=i+1;}stu[i].s[0].c='A';}sort(stu,stu+n,cmp2);for(int i=0;i<n;i++){if(i==0) stu[i].s[1].r=i+1;else if(i>0){if(stu[i].s[1].score==stu[i-1].s[1].score) stu[i].s[1].r=stu[i-1].s[1].r;else stu[i].s[1].r=i+1;}stu[i].s[1].c='C';}sort(stu,stu+n,cmp3);for(int i=0;i<n;i++){if(i==0) stu[i].s[2].r=i+1;else if(i>0){if(stu[i].s[2].score==stu[i-1].s[2].score) stu[i].s[2].r=stu[i-1].s[2].r;else stu[i].s[2].r=i+1;}stu[i].s[2].c='M';}sort(stu,stu+n,cmp4);map<string,int>m,M;//m存id对应的下标，M判断id是否合法for(int i=0;i<n;i++){if(i==0) stu[i].s[3].r=i+1;else if(i>0){if(stu[i].s[3].score==stu[i-1].s[3].score) stu[i].s[3].r=stu[i-1].s[3].r;else stu[i].s[3].r=i+1;}stu[i].s[3].c='N';//我将E先改成M，方便以A-C-M-N的顺序排序比较m[stu[i].id]=i;M[stu[i].id]=1;}string id;while(q--){cin>>id;if(M[id]==0) puts("N/A");else{sort(stu[m[id]].s,stu[m[id]].s+4,cmp5);//比较排名if(stu[m[id]].s[0].c=='N') {stu[m[id]].s[0].c='E';cout<<stu[m[id]].s[0].r<<" "<<stu[m[id]].s[0].c<<endl;}else cout<<stu[m[id]].s[0].r<<" "<<stu[m[id]].s[0].c<<endl;}}
}