1025 PAT Ranking (25分)

文章目录

• 问题
• 解决方法
• 后记

问题

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.
Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (≤300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.
Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.
Sample Input:

2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85

Sample Output:

9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4

解决方法

分析：双重排序。

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct student
{
	string id;
	int grade;
	int finalrank;
	int testlocationrank;
	int testlocation;
};
bool cmp1(student a, student b)
{
	if (a.grade != b.grade) return a.grade > b.grade;
	else return a.id < b.id;
}
int main()
{
	int n, k;
	int cnt = 0, sum = 0;
	scanf("%d", &n);
	vector<struct student>vec(30005);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &k);
		sum += k;
		for (int j = 0; j < k; j++)
		{
			cin >> vec[cnt].id;
			cin.ignore();
			scanf("%d",&vec[cnt].grade);
			vec[cnt].testlocation = i + 1;
			cnt++;
		}
		sort(vec.begin() + cnt - k, vec.begin() + cnt, cmp1);
		vec[cnt - k].testlocationrank = 1;
		for (int j = 1; j < k; j++)
		{
			if (vec[cnt-k+j].grade == vec[cnt-k+j-1].grade) vec[cnt - k + j].testlocationrank = vec[cnt - k + j - 1].testlocationrank;
			else vec[cnt-k+j].testlocationrank = j+1;
		}
	}
	sort(vec.begin(), vec.begin()+sum, cmp1);
	vec[0].finalrank = 1;
	for (int i = 1; i < sum; i++)
	{
		if (vec[i].grade == vec[i - 1].grade) vec[i].finalrank = vec[i - 1].finalrank;
		else vec[i].finalrank = i + 1;
	}
	printf("%d\n", sum);
	for (int i = 0; i < sum; i++)
	{
		cout << vec[i].id;
		printf(" %d %d %d\n", vec[i].finalrank, vec[i].testlocation, vec[i].testlocationrank);
	}
	return 0;
}

后记

这里的ID虽然说是13位，用long long时注意一定要控制13位的格式，避免id 000003输出3的错误。用string就不用控制格式。测试点4主要是考察格式的控制。

来源：CSDN

作者：寻余

链接：https://blog.csdn.net/weixin_43725617/article/details/104629867