1012 The Best Rank （25 分）(排序，sort用法)

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本文介绍了一个学生排名系统的设计与实现，该系统针对计算机科学专业一年级学生的成绩进行评估，通过比较四类排名（C语言编程、数学、英语及平均成绩），确定每位学生的最佳排名，并按优先级顺序输出。

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

本人尝试N小时，一直WA，最后参考柳神代码方可AC。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
struct node{
    int id,best;
    int score[4],rank[4];
}stu[2005];
int exist[1000000],flag=-1;
bool cmp(node a,node b){
    return a.score[flag]>b.score[flag];
}
int main(){
    int n,m,id;
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d %d %d %d",&stu[i].id,&stu[i].score[1],&stu[i].score[2],&stu[i].score[3]);
        stu[i].score[0]=(stu[i].score[1]+stu[i].score[2]+stu[i].score[3])/3.0+0.5;//四舍五入
    }
    for(flag=0;flag<=3;flag++){
        sort(stu,stu+n,cmp);
        stu[0].rank[flag]=1;
        for(int i=1;i<n;i++){
            stu[i].rank[flag]=i+1;
            if(stu[i].score[flag]==stu[i-1].score[flag])
                stu[i].rank[flag]=stu[i-1].rank[flag];
        }
    }
    for(int i=0;i<n;i++){
        exist[stu[i].id]=i+1;//便于最后查找学生id
        stu[i].best=0;
        int minn=stu[i].rank[0];
        for(int j=0;j<=3;j++){
            if(stu[i].rank[j]<minn){
                minn=stu[i].rank[j];
                stu[i].best=j;
            }
        }
    }
    char c[5]={'A','C','M','E'};
    for(int i=0;i<m;i++){
        scanf("%d",&id);
        int temp=exist[id];
        if(temp){
            int best=stu[temp-1].best;
            printf("%d %c\n",stu[temp-1].rank[best],c[best]);
        }else{
            printf("N/A\n");
        }
    }
    return 0;
}

附上sort用法