1025. PAT Ranking (25)

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最新推荐文章于 2022-08-18 21:04:17 发布

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分类专栏：浙大PAT 面试编程训练文章标签： ZJU PAT

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本文深入探讨了使用C++实现结构体排序和合并的技巧，包括如何使用STL库中的容器和算法来简化过程，并详细解释了如何处理相同分数的排名问题。

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考察结构体排序以及合并操作

#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;

typedef struct Node
{
	string registrition_number;
	int grade;
	int location_number;
	int local_rank;
	int final_rank;
	bool operator>(const Node& orh) const
	{
		if(grade != orh.grade)
			return grade>orh.grade;
		else return registrition_number<orh.registrition_number;
	}
}Node;
int main()
{
	int n;
	scanf("%d",&n);
	vector<Node> globalList;
	for(int l = 0; l < n; ++l)
	{
		int k;
		scanf("%d",&k);
		vector<Node> localList(k);
		for(int i = 0; i < k; ++i)
		{
			cin>>localList[i].registrition_number;
			scanf("%d",&localList[i].grade);
			localList[i].location_number = l;
			//globalList.push_back(localList[i]);
		}
		//sort
		sort(localList.begin(), localList.end(), greater<Node>());
		//process the same grade with the same rank
		int nowGrade = -1;
		int nowRank = -1;
		for(int i = 0; i < k; ++i)
		{
			if(localList[i].grade == nowGrade)
				localList[i].local_rank = nowRank;
			else
			{
				nowGrade = localList[i].grade;
				nowRank = i;
				localList[i].local_rank = i;
			}
			globalList.push_back(localList[i]);
		}
	}
	//sort the global list
	//sort
	sort(globalList.begin(), globalList.end(), greater<Node>());
	//process the same grade with the same rank
	int nowGrade = -1;
	int nowRank = -1;
	for(int i = 0; i < globalList.size(); ++i)
	{
		if(globalList[i].grade == nowGrade)
			globalList[i].final_rank = nowRank;
		else
		{
			nowGrade = globalList[i].grade;
			nowRank = i;
			globalList[i].final_rank = i;
		}
	}
	//output
	printf("%d\n", globalList.size());
	for(int i = 0; i < globalList.size(); ++i)
	{
		cout<<globalList[i].registrition_number;
		printf(" %d %d %d\n", globalList[i].final_rank+1, globalList[i].location_number+1, globalList[i].local_rank+1);
	}
	return 0;
}