1025 PAT Ranking (25 分)

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 Nranklists 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

题目大意：有n个考场，每个考场输入k（k为变量）个人数，输入他们的准考证号，分数；将考生分数进行排序，分数相同，按准考证号大小排序，并排出他们的总排名，本考场排名；输出：考生总人数，总排名，考场号，本考场排名

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

struct Student
{
	char id[15];
	int score;
	int location_number;
	int location_rank;
};

bool cmp(Student a,Student b)
{
	if (a.score!=b.score)return a.score>b.score;
	else return strcmp(a.id,b.id)<0;
	//分数相等，则按准考证号大小输出 
} 

int main()
{
	Student stu[30010];
	int n,k,count=0;//count记总考生人数 
	cin>>n;
	for (int i=1; i<=n; i++)
	{
		cin>>k;
		for (int j=0; j<k; j++)
		{
			cin>>stu[count].id>>stu[count].score;
			stu[count].location_number=i;
			count++;
		}
		sort (stu+count-k,stu+count,cmp);//本考场考生排序 
		stu[count-k].location_rank=1;//记下该考场的第一名 
		for (int j=count-k+1; j<count; j++)
		{
			if (stu[j].score==stu[j-1].score)
			{
				stu[j].location_rank=stu[j-1].location_rank;
			}
			else 
			{//在该考场的排名
				stu[j].location_rank=j-(count-k)+1; 
			}//第一名已排出来，所以排名从2开始，
			//j从count-k+1开始，j-(count-k)就是从1开始，+1就是从2开始 
		}
	}
	sort(stu,stu+count,cmp);//所有考生排序 
	cout<<count<<endl;//输出总考生数
	int rank=1;//总排名
	//输出第一名 
	cout<<stu[0].id<<" "<<rank<<" "<<stu[0].location_number<<" "<<stu[0].location_rank<<endl;
	for (int i=1; i<count; i++)
	{
		if (stu[i].score!=stu[i-1].score)
		rank=i+1;//分数不等，排名往后推 
		cout<<stu[i].id<<" "<<rank<<" "<<stu[i].location_number<<" "<<stu[i].location_rank<<endl;	
	} 
	return 0;
}