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

題目大意

有n個考場，每個考場有一定數(shù)量的考生，按照排名、準考證號升序排列，輸出準考證號、分數(shù)、考場號、考場內排名。

思路

主要利用sort函數(shù)完成排序操作，步驟如下：

• 先讀入每個考場的信息，并進行排序；
• 所有考場匯總后再進行排序。

代碼實現(xiàn)

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

struct Student
{
    char id[15];
    int score;
    int trank;      // 總排名
    int lo;         // 考場編號
    int lrank;      // 地區(qū)排名
} stu[30010];

bool cmp(Student a, Student b)
{
    if (a.score != b.score)
        return a.score > b.score;       // 分數(shù)不同，按排名升序排列
    else
        return strcmp(a.id, b.id) < 0;      // 分數(shù)相同，按準考證號升序排列
}

int main(void)
{
    int n;              // 考場數(shù)量
    scanf("%d", &n);
    int num;            // 當前考場考生數(shù)
    int order = 1;      // 記錄當前考生序號
    int total = 0;      // 統(tǒng)計總考生數(shù)
    
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &num);
        for (order; order <= num + total; order++)
        {
            scanf("%s %d", stu[order].id, &stu[order].score);
            stu[order].lo = i;
        }
        sort(stu+total+1, stu+order, cmp);
        stu[total+1].lrank = 1;
        for (int j = total+2; j < order; j++)
            if (stu[j].score == stu[j-1].score)
                stu[j].lrank = stu[j-1].lrank;
            else
                stu[j].lrank = j - total;
        total += num;
    }
    sort(stu+1, stu+order, cmp);
    stu[1].trank = 1;
    for (int i = 2; i <= total; i++)
        if (stu[i].score == stu[i-1].score)
            stu[i].trank = stu[i-1].trank;
        else
            stu[i].trank = i;
    printf("%d\n", total);
    for (int i = 1; i <=total; i++)
        printf("%s %d %d %d\n", stu[i].id, stu[i].trank, stu[i].lo, stu[i].lrank);
}