PAT甲级——（1025） PAT Ranking

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

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
struct student {
	long long no;
	int score, finrank, loca, locarank;//分数、最终排名、考场、考场排名
};
bool cmpl(student a, student b) {
	return a.score != b.score ? a.score > b.score:a.no < b.no;
}
int main() {
	int n, m;
	scanf("%d", &n);
	vector<student>fin;
	for(int i=1;i<=n;i++){
		scanf("%d", &m);
		vector<student>v(m);
		for (int j = 0; j < m; j++) {
			scanf("%lld %d", &v[j].no, &v[j].score);
			v[j].loca = i;
		}
		sort(v.begin(), v.end(), cmpl);
		v[0].locarank = 1;
		fin.push_back(v[0]);
		for (int j = 1; j < m; j++) {
			v[j].locarank = (v[j].score == v[j - 1].score) ? (v[j - 1].locarank) : (j + 1);//相同分数的排名一样
			fin.push_back(v[j]);
		}
	}
	sort(fin.begin(), fin.end(), cmpl);
	fin[0].finrank = 1;
	for (int j = 1; j < fin.size(); j++)
		fin[j].finrank = (fin[j].score == fin[j - 1].score) ? (fin[j - 1].finrank) : (j + 1);
	printf("%d\n", fin.size());
	for (int i = 0; i < fin.size(); i++)
		printf("%013lld %d %d %d\n", fin[i].no, fin[i].finrank, fin[i].loca, fin[i].locarank);
	return 0;	
}

题意：有n个考场，每个考场有若干个数量的考生。现在给出各个考场中考生的准考证号与分数，要求将所有考生按分数从高到低排序，并按顺序输出所有考生的准考证号、排名、考场号以及考场内排名。
思路：
1.按考场读入各考生的信息，并对当前读入考场的所有考生进行排序。之后将该考场的所有考生的排名写入他们的结构体内。
2.对所有考生进行排序。
3.按顺序一边计算总排名，一边输出所有考生的信息。
注意：相同的分数情况下按照学号的从小到大排序，但是他们的排民应该是一样的数字。