1141 PAT Ranking of Institutions (25 point(s))

PAT竞赛机构排名算法解析

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本文详细介绍了一种用于PAT竞赛的机构排名算法，该算法基于参赛者的成绩进行加权计算，考虑了不同级别考试的成绩差异，并通过总加权分数、参赛者数量及机构代码的字母顺序来确定最终排名。

1141 PAT Ranking of Institutions (25 point(s))

After each PAT, the PAT Center will announce the ranking of institutions based on their students' performances. Now you are asked to generate the ranklist.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤105), which is the number of testees. Then N lines follow, each gives the information of a testee in the following format:

ID Score School

where ID is a string of 6 characters with the first one representing the test level: B stands for the basic level, Athe advanced level and T the top level; Score is an integer in [0, 100]; and School is the institution code which is a string of no more than 6 English letters (case insensitive). Note: it is guaranteed that ID is unique for each testee.

Output Specification:

For each case, first print in a line the total number of institutions. Then output the ranklist of institutions in nondecreasing order of their ranks in the following format:

Rank School TWS Ns

where Rank is the rank (start from 1) of the institution; School is the institution code (all in lower case); ; TWS is the total weighted score which is defined to be the integer part of ScoreB/1.5 + ScoreA + ScoreT*1.5, where ScoreX is the total score of the testees belong to this institution on level X; and Ns is the total number of testees who belong to this institution.

The institutions are ranked according to their TWS. If there is a tie, the institutions are supposed to have the same rank, and they shall be printed in ascending order of Ns. If there is still a tie, they shall be printed in alphabetical order of their codes.

Sample Input:

10
A57908 85 Au
B57908 54 LanX
A37487 60 au
T28374 67 CMU
T32486 24 hypu
A66734 92 cmu
B76378 71 AU
A47780 45 lanx
A72809 100 pku
A03274 45 hypu

Sample Output:

5
1 cmu 192 2
1 au 192 3
3 pku 100 1
4 hypu 81 2
4 lanx 81 2

中规中矩的模拟题。

考察点：重载运算符、STL的使用。

注意：

1. 将字符串变成小写：transform(s.begin(),s.end(),s.begin(),::tolower);

2. 名次赋值；

3. 迭代器的使用。

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<ctype.h>
#include<map>
using namespace std;
struct InsRec{
	string ins;int ns;int tws;int rank;
	InsRec(string i,int n,int t):ins(i),ns(n),tws(t){}
	bool operator < (const InsRec &other) const{
		if(tws!=other.tws) return tws>other.tws;
		if(ns!=other.ns) return ns<other.ns;
		return ins<other.ins;
	}
};
map<string,vector<double> >mp;
vector<InsRec> v;
int main(void){
	int N;cin>>N;
	string id,ins;double grade;
	for(int i=0;i<N;i++){
		cin>>id>>grade>>ins;
		if(id[0]=='B') grade = grade/1.5;
		if(id[0]=='T') grade = grade*1.5;
		transform(ins.begin(),ins.end(),ins.begin(),::tolower);
		mp[ins].push_back(grade);
	}
	map<string,vector<double> >::iterator it;
	for(it = mp.begin();it!=mp.end();it++){
		double sumGrade=0.0;
		for(int i=0;i<(int)(it->second).size();i++){
			sumGrade+=(it->second)[i];
		}
		int cnt = (int)(it->second).size();
		v.push_back(InsRec((it->first),cnt,(int)sumGrade));
	}
	sort(v.begin(),v.end());
	for(int i=0;i<v.size();i++){
		if(i==0||(i>0&&v[i].tws!=v[i-1].tws)) v[i].rank=i+1;
		else v[i].rank = v[i-1].rank;
	}
	cout<<v.size()<<endl;
	for(int i=0;i<v.size();i++){
		cout<<v[i].rank<<" "<<v[i].ins<<" "<<v[i].tws<<" "<<v[i].ns<<endl;
	}
	return 0;
}