PAT 1025 PAT Ranking [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

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

我的代码：

分组排序，再整体排序。sort函数可自定义排序的起点和终点。

#include<cstdio>
#include<algorithm>
#include<string.h>

using namespace std;

struct student
{
	char sno[30]; //学号 
	int score; //分数 
	int rankNow,rank; //最终排名和分组排名 
	int location;  //所属小组 
}stu[30005]; // N<=100 K<=300 

bool cmp(student a,student b)
{
	if(a.score!=b.score) return a.score>b.score;
	//字符串不能直接比较，需借助strcmp函数，小于零说明a.sno<b.sno 
	return strcmp(a.sno,b.sno)<0; //分数相同，学号小在前大在后 
}

int main()
{
	int n,i,j,count = 0;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		int x;
		scanf("%d",&x);
		for(j=0;j<x;j++)
		{
			scanf("%s%d",&stu[count].sno,&stu[count].score);
			stu[count].location = i+1;
			count++;
		}
		//起首元素和尾元素分别是不同小组的起点和终点 
		sort(stu+count-x,stu+count,cmp);
		//排序完成，数组第一个元素排名为1 
		stu[count-x].rank = 1;
		for(j=count-x+1;j<count;j++)
		{
			//如果分数相等，则排名相等 
			if(stu[j].score==stu[j-1].score) 
				stu[j].rank = stu[j-1].rank;
			//否则等于数组下标+1，-(count-x)还是因为分组的缘故 
			else stu[j].rank = j+1-(count-x);
		}
	}
	//记录完分组排名后进行整体排序 
	sort(stu,stu+count,cmp);
	printf("%d\n",count);
	int rank = 1;
	for(i=0;i<count;i++)
	{
		if(i>0&&stu[i].score==stu[i-1].score) 
			stu[i].rankNow = stu[i-1].rankNow;
		else rank = i+1;
		stu[i].rankNow = rank;
	}
	for(i=0;i<count;i++)
		printf("%s %d %d %d\n",stu[i].sno,stu[i].rankNow,stu[i].location,stu[i].rank);
}