PAT 1012 The Best Rank [排序]

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该博客讨论了PAT 1012题目的解决方案，涉及对一年级CS学生在C语言、数学和英语三门课程成绩的评估。文章解释了如何根据学生的最好排名来鼓励他们，特别强调了排名的优先级：C > M > E > 平均分。博主提供了输入输出规格，并给出了处理学生最佳排名的策略，包括处理排名相同的情况和按优先级输出。此外，还包含了样例输入和输出以及博主的代码实现思路。

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 Nlines 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

我的代码：

个人最好名次的科目和排名→每门科目的排名→单科成绩排序并记录名次*4（包括平均分4门科目循环4次）

排名相同时按照ACME的优先级别输出↓

结构体中开数组按照ACME的顺序分别记录成绩和排名，遍历时让记录最小排名的"指针"min的初始值为0

（只有大于min的排名才可替换min，这样可保证相等时按照优先顺序输出）

#include<cstdio>
#include<algorithm>
#include<string.h>

using namespace std;

struct node
{
	char id[10]; //才6位数也可以用int类型 
	int score[4],rank[4];
}stu[2008];

//需定义全局变量供cmp调用 
int num;

bool cmp(node a,node b)
{
	return a.score[num]>b.score[num];
}

int main()
{
	int i,j,n,m;
	char sub[4] = {'A','C','M','E'};
	scanf("%d%d",&n,&m);
	for(i=0;i<n;i++)
	{
		scanf("%s%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];
		//平均分相加即可比较，不必过分计较 
	}
	//四门科目，四次排序，记录四个排名情况 
	for(num=0;num<4;num++)
	{
		sort(stu,stu+n,cmp);
		stu[0].rank[num] = 1;
		for(i=1;i<n;i++)
		{
			if(stu[i].score[num]!=stu[i-1].score[num]) stu[i].rank[num] = i+1;
			else stu[i].rank[num] = stu[i-1].rank[num];
		}
	}	
	for(i=0;i<m;i++)
	{
		char query[10];
		int flag = 0;
		scanf("%s",&query);
		for(j=0;j<n;j++)
		{
			if(strcmp(query,stu[j].id)==0)
			{
				flag = 1;
				break;
			}
		}
		if(flag)
		{
			int min = 0;
			for(int k=0;k<4;k++)
			{
				//＜不可以<=，否则无法优先顺序输出 
				if(stu[j].rank[k]<stu[j].rank[min]) min = k; 		
			}
			printf("%d %c\n",stu[j].rank[min],sub[min]);
		}			
		else printf("N/A\n");
	}
}