PAT 甲级 1025 1025 PAT Ranking (25 分) STL +sort 卡在最后一个测试点

偏执的太偏执、 2024-02-19 21:28 55阅读 0赞

## [1025 PAT Ranking (25 分)][1025 PAT Ranking _25] ##

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

这题很实际，每个考场都有自己的考场号、自己的考场排名，还有全局排名

但是我卡了两个点：都是因为题意没读太懂，一个是第二个测试点，没有发现：如果分数相同，按注册号非递减顺序输出。

还有一个是第四个测试点，注册号也就是id，最好声明成string类型，这样就避免了 ID是 000003 输出 ：3的 小错误。

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    struct node{
    	string id;
    	int grade,f_r,ln,lr;
    };
    int cmp(node a,node b)
    {
    	return a.grade>b.grade;
    }
    int cmp1(node a,node b)
    {
    	if(a.grade==b.grade)
    	return a.id<b.id;
    	else
    	return a.grade>b.grade;
    }
    int main()
    {
    	vector<node> a[103];
    	vector<node> b;
    	node t;
    	int n,m,i,j;
    	cin>>n;
    	for(i=0;i<n;i++)
    	{
    		cin>>m;
    		if(m==0)
    		continue;
    		for(j=0;j<m;j++)
    		{
    			cin>>t.id;
    			cin>>t.grade;
    			t.ln=i+1;
    			a[i].push_back(t);
    		}
    	}
    	for(i=0;i<n;i++)
    	{
    		if(a[i].size()==0)
    		continue;
    		sort(a[i].begin(),a[i].end(),cmp);
    		a[i][0].lr=1;
    		b.push_back(a[i][0]);
    		for(j=1;j<a[i].size();j++)
    		{
    			if(a[i][j-1].grade==a[i][j].grade)
    			a[i][j].lr=a[i][j-1].lr;
    			else
    			a[i][j].lr=j+1;
    			b.push_back(a[i][j]);
    		}
    	}
    	if(b.size()==0)
    	return 0;
    	sort(b.begin(),b.end(),cmp1);
    	b[0].f_r=1;
    	for(i=1;i<b.size();i++)
    	{
    		if(b[i].grade==b[i-1].grade)
    		b[i].f_r=b[i-1].f_r;
    		else
    		b[i].f_r=i+1;
    	}
    	cout<<b.size()<<endl;
    	for(i=0;i<b.size();i++)
    	cout<<b[i].id<<" "<<b[i].f_r<<" "<<b[i].ln<<" "<<b[i].lr<<endl;
    	return 0;
     }

[1025 PAT Ranking _25]: https://pintia.cn/problem-sets/994805342720868352/problems/994805474338127872