PAT（甲级）1141.PAT Ranking of Institutions （25 分)
PAT（甲级）1141.PAT Ranking of Institutions （25 分)
PAT 1141.PAT Ranking of Institutions （25 分)
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.
输入格式：
Each input file contains one test case. For each case, the first line gives a positive integer N (≤10^5), 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,A the advanced level andTthe 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.
输出格式：
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.
输入样例：
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输出样例：
5
1 cmu 192 2
1 au 192 3
3 pku 100 1
4 hypu 81 2
4 lanx 81 2题目分析：sort的使用，测试点五是一个坑，因为每个学校的分数相当于一个加权的成绩，在前期处理的时候就应该按照浮点数处理，只有在排序的时候将其转换为整数即可。
其中使用C++11的unordered_map和unordered_set会大大提高效率，因为其底层是使用hash_table实现的，具体可以参阅C++11的新标准。
AC代码：
#include <iostream>
#include <string>
#include <vector>
#include <unordered_map>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 100010;
struct school{
string name;
int score;
int sum;
school(){}
school(string name, int score, int sum):name(name), score(score), sum(sum){}
};
vector<school> v;
unordered_map<string, double> nameToScore;
unordered_map<string, int> nameToSumPeople;
set<string> schoolName;
void toLower(string& s){
for(int i=0; i<s.length(); ++i){
if(s[i]>='A' && s[i]<='Z')s[i] += 32;
}
}
bool cmp(school a, school b){
if(a.score!=b.score)return a.score>b.score;
if(a.sum!=b.sum)
return a.sum < b.sum;
else
return a.name < b.name;
}
int main(){
//freopen("in.txt", "r", stdin);
int n;
scanf("%d", &n);
string id, name;
double score;
for(int i=0; i<n; ++i){
cin>>id;
scanf("%lf", &score);
cin>>name;
toLower(name);
schoolName.insert(name);
nameToSumPeople[name] += 1;
if('A'==id[0]){nameToScore[name] += score;}
else if('B'==id[0]){nameToScore[name] += (score / 1.5);}
else{nameToScore[name] += (score * 1.5);}
}
for(auto it=schoolName.begin(); it!=schoolName.end(); ++it){
v.push_back(school(*it, (int)(nameToScore[*it]), nameToSumPeople[*it]));
}
sort(v.begin(), v.end(), cmp);
cout<<v.size()<<endl;
int rank = 1;
cout<<rank<<" "<<v[0].name<<" "<<v[0].score<<" "<<v[0].sum<<endl;
for(int i=1; i<v.size(); ++i){
if(!i || v[i].score!=v[i-1].score)
rank = i+1;
cout<<rank<<" "<<v[i].name<<" "<<v[i].score<<" "<<v[i].sum<<endl;
}
return 0;
}腾讯云开发者
