【笨方法学PAT】1012 The Best Rank（25 分）

一、题目

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 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

二、题目大意

给定学生的C、M、E三科成绩，求C、M、E、平均成绩A中的最好名次。

三、考点

排序

四、解题思路

这道题很繁琐，需要排序的东西太多了：ACME四个成绩，还要分别求对应的Rank。如果分开求，代码会很长，使用struct嵌套数组，可以减小工作量；

1、使用 struct 保存数据，按照ACME存储成绩；

2、使用 cmp_flag 全局变量保存排序的科目，排序完成之后，求 Rank ，成绩相同排名相同，否则与当前的位置相同；

3、之后判断的时候，先判断 ID 是否存在，存在的话，循环4个项目，获得最小排名。

五、代码

#include<iostream>
#include<algorithm>
#include<vector>
#define N 1000010
#define INF 99999999
using namespace std;
struct node {
	int id;
	//a c m e
	int g[4],rank[4];
};
int cmp_flag = 0;
bool cmp(node n1, node n2) {
	return n1.g[cmp_flag] > n2.g[cmp_flag];
}
int main() {
	int n, m;
	cin >> n >> m;
	vector<node> vec(n);
	bool show[N] = { false };

	//读取数据
	for (int i = 0; i < n; ++i) {
		cin >> vec[i].id >> vec[i].g[1] >> vec[i].g[2] >> vec[i].g[3];
		vec[i].g[0] = (vec[i].g[1] + vec[i].g[2] + vec[i].g[3]) / 3;
		show[vec[i].id] = true;
	}

	//排名
	for (int i = 0; i < 4; ++i) {
		cmp_flag = i;
		sort(vec.begin(), vec.end(), cmp);

		vec[0].rank[i] = 1;
		for (int j = 1; j < vec.size(); ++j) {
			//成绩相等，排名相同
			if (vec[j].g[i] == vec[j - 1].g[i]) {
				vec[j].rank[i] = vec[j - 1].rank[i];
			}
			//成绩不相等，排名靠后
			else
				vec[j].rank[i] = j + 1;
		}
	}

	while (m--) {
		int id;
		cin >> id;

		//没有找到
		if (show[id]==false) {
			cout << "N/A"<<endl;
			continue;
		}

		//排名
		int best_rank = INF;
		int best_id = 0;
		char int2char[4] = { 'A','C','M','E' };

		//排序之后顺序改变，需要找到 id 的位置
		node tmp;
		for (int i = 0; i < vec.size(); ++i) 
			if (id == vec[i].id) {
				tmp = vec[i];
				break;
			}
		//寻找最小rank
		for (int i = 0; i < 4; ++i) {
			if (tmp.rank[i] < best_rank) {
				best_rank = tmp.rank[i];
				best_id = i;
			}
		}
		cout << best_rank << " " << int2char[best_id] << endl;
	}

	system("pause");
	return 0;
}