[ PAT-A ] 1025 PAT Ranking (C++)

题目描述

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
给定多组参赛人员编号和其对应的分数,求出参赛人员在全部参赛人员中的排名,以及在其对应组组内的排名情况
思路
水题,但要注意排序的”姿势”

代码设计

//结构体和全局变量
//zhicheng
typedef struct{string num;int sco;int fin_rank;int flg;int loc_rank;}D;
vector<D> re;//存放全体参赛人员信息
vector<D> te; //存放每组参赛人员信息

[任务] 对一个数组进行排序,可能是全局排序,或者是组内排序
[接口]
inline void _sort(vector<D>& a,int fg)
输入:a 含参赛人员信息的数组 ,fg表示进行所在组组内排序还是全体人员排序(1:全体,0:组内)

//代码实现
//zhicheng
inline void _sort(vector<D>& a,int fg)
{
    sort(a.begin(),a.end(),comp);
    if(!fg) {a[0].loc_rank=1; re.push_back(a[0]);}
    else a[0].fin_rank=1;
    for(int i=1;i<a.size();i++)
    {
        if(!fg)
        {
            a[i].loc_rank=(a[i].sco==a[i-1].sco)?(a[i-1].loc_rank):(i+1);
            re.push_back(a[i]);
        }
        else 
        {
            a[i].fin_rank=(a[i].sco==a[i-1].sco)?(a[i-1].fin_rank):(i+1);
        }
    }
}

有关PAT (Basic Level) 的更多内容可以关注 ——> PAT-B题解

有关PAT (Advanced Level) 的更多内容可以关注 ——> PAT-A题解

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[ PAT-A ] 1025 PAT Ranking (C++)

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