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
题目大意：给出每个学生的id、分数、学校，学校名称不区分大小写，输出学校排名、学校名称、总加权成绩、学校参赛人数。学校名称输出时候以小写方式输出。
分析：两个map，一个cnt用来存储某学校名称对应的参赛人数，另一个sum计算某学校名称对应的总加权成绩。每次学校名称string school都要转化为全小写，将map中所有学校都保存在vector ans中，类型为node，node中包括学校姓名、加权总分、参赛人数。对ans数组排序，根据题目要求写好cmp函数，最后按要求输出。对于排名的处理：设立pres表示前一个学校的加权总分，如果pres和当前学校的加权总分不同，说明rank等于数组下标+1，否则rank不变～
注意：总加权分数取整数部分是要对最后的总和取整数部分，不能每次都直接用int存储，不然会有一个3分测试点不通过～
PS：之前直接使用map，导致在新的PAT系统中提交后最后一个测试点超时，改成了unordered_map即可AC～
#include <iostream>
#include <algorithm>
#include <cctype>
#include <vector>
#include <unordered_map>
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;
cin >> id;
double score;
scanf("%lf", &score);
cin >> 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 = 0, pres = -1;
printf("%d\n", (int)ans.size());
for (int i = 0; i < ans.size(); i++) {
if (pres != ans[i].tws) rank = i + 1;
pres = ans[i].tws;
printf("%d ", rank);
cout << ans[i].school;
printf(" %d %d\n", ans[i].tws, ans[i].ns);
}
return 0;
}