1025. PAT Ranking (25) [排序]

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
1
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

//先对同一考场考生排序排名，再对全体考生排序排名 
#include<cstdio> 
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

//定义学生结构体 
struct Student{
	char registration_number[15]; //准考证号 
	int total_score; //总分 
	int location_number; //考场号 
	int local_rank; //考场内排名 
}stu[30010];	//注意这里的取值 **题目规定不超过100个考场，每个考场不超过300个人，所以该值不小于100×300=30000人。  

//定义比较函数 
bool cmp(Student a, Student b){
	if (a.total_score!=b.total_score) return a.total_score > b.total_score; 
	else return strcmp(a.registration_number,b.registration_number)<0; 
}

int main(){
	int n, k, num=0; // n 考场数; k 当前考场人数; num 总考生数
	scanf("%d",&n); 
	
	//分考场输入考生信息 
	for(int i=1; i<=n; i++){
		//该考场人数 
		scanf("%d",&k); 
		//输入每位考生的准考证号，总分，考场号
		for(int j=0;j<k;j++){
			scanf("%s %d",stu[num].registration_number, &stu[num].total_score); 
			stu[num].location_number=i; 
			num++; //总考生数 ++ 
		}	
		//考场内排序
		sort(stu+num-k, stu+num, cmp);	//num-k即（5-5或9-4）保证两个考场都适用
		//考场内排名
		stu[num-k].local_rank=1; 
		for(int j=num-k+1; j<num; j++){
			if(stu[j].total_score==stu[j-1].total_score){ //与前一名总分相同，前一名的名次给后一名 
				stu[j].local_rank = stu[j-1].local_rank;
			}
			else{		//否则，后一名的排名+1（num-k使得两个考场都适用）
				stu[j].local_rank = j+1-(num-k);
			}
		}
	}

	//输出总考生数 
	printf("%d\n",num);
	
	//对所有考生排序
	sort(stu, stu+num, cmp) ;
	
	//对所有考生排名;输出每位学生的准考证号，总排名，考场号，考场排名 
	int final_rank=1;
	for(int i=0; i<num; i++){
		if(i>0 && stu[i].total_score!=stu[i-1].total_score){
			final_rank=i+1;
		}
		printf("%s %d %d %d\n",stu[i].registration_number,final_rank,stu[i].location_number,stu[i].local_rank);
	}
	
	return 0;
}