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本文介绍了一个用于合并多个考试地点的编程能力测试(PAT)成绩的程序。该程序能够正确地合并所有成绩，并按最终排名输出结果。输入包括多个考试地点的成绩列表，每个列表包含考生的准考证号和分数。输出则按照最终排名、所在考场和考场内排名显示所有考生的成绩。

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
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct Student
{
	char id[15];//准考证号 
	int score;//分数
	int location_number;//考场号
	int local_rank;//考场内排名 
}stu[30010];
bool cmp(Student a,Student b)
{
	if(a.score!=b.score)
	{
		return a.score>b.score;//分数不相同的时候，从高到低排序 
	}
	else 
	{
		return strcmp(a.id,b.id)<0;
	} 
}

int main()
{
	int n,k,num=0;//num为总考场数
	scanf("%d",&n);
	for(int i=1;i<=n;i++) //i代表考场号 
	{
		scanf("%d",&k);  //K代表考场里面有几个学生 
		for(int j=0;j<k;j++)
		{
			scanf("%s %d",stu[num].id,&stu[num].score);
			stu[num].location_number=i;
			num++;
		}
		sort(stu+num-k,stu+num,cmp);//该考场内部学生排序 
		stu[num-k].local_rank=1;//该考场第一名的local_rank记为1
		
		for(int j=num-k+1;j<num;j++)
		{
			if(stu[j].score==stu[j-1].score)
			{
				//如果分数相同，则排名相同 
				stu[j].local_rank=stu[j-1].local_rank;
			}
			else//如果不同分 
			{
				//local_rank为该考生前面的人数 
				stu[j].local_rank=j+1-(num-k);
			} 
		}
	}
	cout<<num<<endl; 
	sort(stu,stu+num,cmp);//将所有考生排序
	int r=1;
	
	for(int i=0;i<num;i++)
	{
		if(i>0&&stu[i].score!=stu[i-1].score)
		{
			r=i+1; //当前考生与上一个考生分数不同时，
			//让r更新为人数+1; 
		}
		
		printf("%s ",stu[i].id);
		
		cout<<r<<" "<<stu[i].location_number<<" "<<stu[i].local_rank<<endl;
	} 

	return 0;
	
	 
}