(PAT 1141) PAT Ranking of Institutions (排序+unorded_map的使用)

我不是女神ヾ 2022-03-21 14:54 180阅读 0赞

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, `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

解题思路:

利用map的映射建立对应关系,由题目涉及到排序,而map中无法根据value进行排序,所以我们在读完数据计算完总成绩后,需要把map中的键值对放到一个vector中,然后在vector中根据键值对的值进行排序

这里如果使用map的话,最后两个测试点会超时.改用unorded\_map就可以解决问题

unorded\_map是无序的map,其实现原理是哈希表映射,查找效率为O(1)

这篇博客中博主很好地说明了map和unorded\_map的关系: [https://blog.csdn.net/BillCYJ/article/details/78985895][https_blog.csdn.net_BillCYJ_article_details_78985895]

该题也要注意int和double转换的问题,输入和计算时是double型,输出时为int型

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <unordered_map>
    #include <set>
    #include <string>
    #include <string.h>
    using namespace std;
    unordered_map<string, double>  schoolToScore;
    unordered_map<string, int> schoolToPeople;
    vector<pair<string, int>> mainMap;
    bool cmp(pair<string, int>& a, pair<string, int>& b) {
    	if (a.second != b.second) return a.second > b.second;
    	else {
    		if (schoolToPeople[a.first] != schoolToPeople[b.first]) return schoolToPeople[a.first] < schoolToPeople[b.first];
    		else {
    			return strcmp(a.first.c_str(), b.first.c_str()) < 0;
    		}
    	}
    }
    int main() {
    	int N;
    	scanf("%d", &N);
    	for (int i = 0; i < N; ++i) {
    		string ID, School;
    		double score;
    		cin >> ID;
    		scanf("%lf", &score);
    		cin >> School;
    		for (int j = 0; j < School.size(); ++j) {
    			School[j] = tolower(School[j]);
    		}
    		
    		if (ID[0] == 'T') {
    			if (!schoolToScore[School]) schoolToScore[School] = score * 1.5;
    			else schoolToScore[School] += score * 1.5;
    		}
    		else if (ID[0] == 'A') {
    			if (!schoolToScore[School]) schoolToScore[School] = score;
    			else schoolToScore[School] += score;
    		}
    		else {
    			if (!schoolToScore[School]) schoolToScore[School] = score / 1.5;
    			else schoolToScore[School] += score / 1.5;
    		}
    		if (!schoolToScore[School]) schoolToPeople[School] = 1;
    		else schoolToPeople[School]++;
    	}
    	
    	for (auto itr = schoolToScore.begin(); itr != schoolToScore.end();++itr) {
    		mainMap.push_back(make_pair(itr->first, (int)itr->second));
    	}
    	sort(mainMap.begin(), mainMap.end(), cmp);
    	
    	printf("%d\n", schoolToPeople.size());
    	int norder = 1;
    	int order = 1;
    	int pres = -1;
    	for (int i = 0; i < mainMap.size(); ++i) {
    		if (pres != mainMap[i].second) {
    			norder = order;
    			printf("%d %s %d %d\n", norder, mainMap[i].first.c_str(), mainMap[i].second, schoolToPeople[mainMap[i].first]);
    			pres = mainMap[i].second;
    			order++;
    		}
    		else {
    			printf("%d %s %d %d\n", norder, mainMap[i].first.c_str(), mainMap[i].second, schoolToPeople[mainMap[i].first]);
    			order++;
    		}
    	}
    	system("PAUSE");
    	return 0;
    }

[https_blog.csdn.net_BillCYJ_article_details_78985895]: https://blog.csdn.net/BillCYJ/article/details/78985895