1012. The Best Rank (25)

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
PAT很喜欢考这种排序题，虽然不是很难，但是很繁琐，代码量一般会长些，需要注意要记录哪些信息，对数组的使用有什么变换，很多情况下都是使用map来映射到数组的下标
#include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
/* 此方法使用map映射id到数组下标
 */
struct Stu
{
	int id;
	int grade[4]; //a, c, m, e及其各自排名
	int best, bestIndex; //最好成绩，及最好成绩对应的科目0：A，1：C；2：M；3：E；
};
vector<Stu> stu;
map<int, int> mp; //id —— index
bool cmp0(Stu a, Stu b)
{
	return a.grade[0] > b.grade[0];
}
bool cmp1(Stu a, Stu b)
{
	return a.grade[1] > b.grade[1];
}
bool cmp2(Stu a, Stu b)
{
	return a.grade[2] > b.grade[2];
}
bool cmp3(Stu a, Stu b)
{
	return a.grade[3] > b.grade[3];
}
bool cmp(Stu a, Stu b)
{
	return a.id < b.id;
}
void getSort(int grade, int n)
{
	switch(grade)
	{
	case 0:sort(stu.begin(), stu.end(), cmp0);break;
	case 1:sort(stu.begin(), stu.end(), cmp1);break;
	case 2:sort(stu.begin(), stu.end(), cmp2);break;
	case 3:sort(stu.begin(), stu.end(), cmp3);break;
	}
	if (stu[0].best > 1)
	{
		stu[0].best = 1;
		stu[0].bestIndex = grade;
	}
	int rank = 1; 
	for (int i = 1; i < n; ++i)
	{
		if (stu[i].grade[grade] != stu[i-1].grade[grade])
			rank = i + 1;
		if (stu[i].best > rank)
		{
			stu[i].best = rank;
			stu[i].bestIndex = grade;
		}
	}
}
int main(void)
{
	int n, m, id;
	scanf("%d%d", &n, &m); \\n和m<=2000
	stu.resize(n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d", &id);
		mp[id] = i;
		scanf("%d%d%d", &stu[i].grade[1], &stu[i].grade[2], &stu[i].grade[3]);
		stu[i].grade[0] = (int)((stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3]) / 3.0 + 0.5); //这里也可以直接等于三个成绩之和，经测试好像都能AC
		stu[i].id = i;    //必须要记住它之前所处的位置，需要靠id值把所有元素重新排好序，好使map映射是正确的
		stu[i].best = n + 1;
	}
	for (int j = 0; j < 4; ++j)
		getSort(j, n);
	sort(stu.begin(), stu.end(), cmp);
	char Ch[4] = {'A', 'C', 'M', 'E'};
	for (int i = 0; i < m; ++i)
	{
		scanf("%d", &id);
		map<int,int>::iterator iter = mp.find(id);
		if (iter == mp.end())
			printf("N/A\n");
		else
		{
			id = iter -> second;
			printf("%d %c\n", stu[id].best, Ch[stu[id].bestIndex]);
		}
	}
	return 0;
}