PAT甲级2017冬季 7-2 1141 PAT Ranking of Institutions (25)

最新推荐文章于 2022-12-02 22:14:11 发布

原创最新推荐文章于 2022-12-02 22:14:11 发布 · 147 阅读

0 ·

CC 4.0 BY-SA版权

PAT算法笔记专栏收录该内容

61 篇文章

订阅专栏

本文介绍了一个用于计算PAT考试中各机构排名的算法。通过综合考虑考生的分数和等级，算法生成了一个基于加权总分（TWS）的排名列表，其中TWS由不同级别分数加权得出。文章详细解释了输入输出规范，以及如何处理同分情况下的排名和机构代码的字母排序。

算法笔记总目录
 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.

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, A the 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

代码：

#include <bits/stdc++.h>
using namespace std;
struct node {
    string school;
    int tws, ns;
};
bool cmp(node a, node b) {
    if (a.tws != b.tws) return a.tws > b.tws;
    else if (a.ns != b.ns) return a.ns < b.ns;
    else return a.school < b.school;
}
int main() {
    int n;
    scanf("%d", &n);
    unordered_map<string, int> cnt;
    unordered_map<string, double> sum;
    for (int i = 0; i < n; i++) {
        string id, school;
        double score;
        cin >>id>>score>>school;
        for (int j = 0; j < school.length(); j++) school[j] = tolower(school[j]);
        if (id[0] == 'B') score = score / 1.5;
        else if (id[0] == 'T') score = score * 1.5;
        sum[school] += score;
        cnt[school]++;
    }
    vector<node> ans;
    for (auto it = cnt.begin(); it != cnt.end(); it++)
        ans.push_back(node{it->first, (int)sum[it->first], cnt[it->first]});
    sort(ans.begin(),ans.end(),cmp);
    int rank = 1;
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < ans.size(); i++) {
        if (i>0 && ans[i-1].tws != ans[i].tws) rank = i + 1;
        printf("%d %s %d %d\n",rank,ans[i].school.c_str(),ans[i].tws, ans[i].ns);
    }
    return 0;
}