1025 PAT Ranking （25 分）多重排序

题目

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

解题思路

　　题目大意： 有N个区域，每个区域有K个学生，现在对这些学生进行排序，一个是该生在所有学生中的排序，一个是该生在该区域的局部排序。
　　解题思路： 分别做两次排序即可，使用vector和sort算法会提高效率。

/*
** @Brief:No.1014 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-11-27 
** @status: Accepted
*/
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cstring>

using namespace std;

struct student{
	string name;
	int score;
	int localNumber;
	int localRank;
	int rank;
};

bool cmp(student&a,student&b){
	if(a.score!=b.score){
		return a.score>b.score;
	}else{
		return a.name<b.name;	
	}
}

int main(){
	vector<student> allStudents;	
	vector<student> local;
	int N,K; 
	while(cin>>N){
		int sumStudents = 0;
		for(int i=1;i<=N;i++){
			cin>>K;
			sumStudents+=K;
			local.clear();
			// input i-th local student.
			for(int j=0;j<K;j++){
				student temp;
				cin>>temp.name>>temp.score;
				temp.localNumber = i;
				local.push_back(temp);
			}
			sort(local.begin(),local.end(),cmp);
			int localRank = 1;
			for(int index = 0;index<K-1;index++){
				local[index].localRank = localRank;
				if(local[index].score!=local[index+1].score){
					localRank = index+2;
				}
			}
			local[K-1].localRank = localRank; 
			allStudents.insert(allStudents.end(),local.begin(),local.end());
		}
		sort(allStudents.begin(),allStudents.end(),cmp);
		int rank = 1;
		for(int index = 0;index<sumStudents-1;index++){
			allStudents[index].rank = rank;
			if(allStudents[index].score!=allStudents[index+1].score){
				rank = index+2;
			}
		}
		allStudents[sumStudents-1].rank = rank;
		cout<<sumStudents<<endl;
		for(auto elem:allStudents){
			cout<<elem.name<<" "<<elem.rank<<" "<<elem.localNumber<<" "<<elem.localRank<<endl; 
		}
	}
	return 0;
}