1153 Decode Registration Card of PAT (25 分)

首先是鬼鬼珊的闲聊一分钟
在之前的博客里，如果你看过的话，一定知道我肯定每天一问，什么时候甜甜的恋爱才能轮到我呀？哈哈哈，然后呢，现在真的轮到我了。是的，超级无敌酷的鬼鬼珊同学真的谈恋爱了，暂时还不知道甜不甜，体会还不深，以后体会深刻的话，一定在博客里认真仔细的写下来。（话就先放在这里，以后肯定写篇博客关于恋爱甜不甜的感受进行深刻的剖析），说一下我的感受吧，大概是第一次谈恋爱的原因，总觉得很恍惚，很不真实，诶，我真的有了男朋友啦？？？最近博主很倒霉，把脚崴了，第一次肿成了猪蹄，以前生病自己都是一个人去看病，早已练就了一身无坚不摧的身心，哈哈哈，略显夸张。但这次真的感受到有男朋友后的不一样体验呢。打住，这不是一篇秀恩爱的博客，这是一篇关于学习的博客，博主的梦想是靠着学习涨粉呢。。。。
言归正传

A registration card number of PAT consists of 4 parts:

the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic;
the 2nd - 4th digits are the test site number, ranged from 101 to 999;
the 5th - 10th digits give the test date, in the form of yymmdd;
finally the 11th - 13th digits are the testee’s number, ranged from 000 to 999.
Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries.

Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤10的4次方 ) and M (≤100), the numbers of cards and the queries, respectively.Then N lines follow, each gives a card number and the owner’s score (integer in [0,100]), separated by a space.After the info of testees, there are M lines, each gives a query in the format Type Term, where
Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term will be the letter which specifies the level;
Type being 2 means to output the total number of testees together with their total scores in a given site. The corresponding Term will then be the site number;
Type being 3 means to output the total number of testees of every site for a given test date. The corresponding Term will then be the date, given in the same format as in the registration card.
Output Specification:
For each query, first print in a line Case #: input, where # is the index of the query case, starting from 1; and input is a copy of the corresponding input query. Then output as requested:

for a type 1 query, the output format is the same as in input, that is, CardNumber Score. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed);
for a type 2 query, output in the format Nt Ns where Nt is the total number of testees and Ns is their total score;
for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt’s, or in increasing order of site numbers if there is a tie of Nt.
If the result of a query is empty, simply print NA.

Sample Input:

8 4
B123180908127 99
B102180908003 86
A112180318002 98
T107150310127 62
A107180908108 100
T123180908010 78
B112160918035 88
A107180908021 98
1 A
2 107
3 180908
2 999

Sample Output:

Case 1: 1 A
A107180908108 100
A107180908021 98
A112180318002 98
Case 2: 2 107
3 260
Case 3: 3 180908
107 2
123 2
102 1
Case 4: 2 999
NA

思路：这个题呢，也没有使用到什么特殊的算法，有可能是我的水平问题，我做这道题的时候，关于解题思路就很直白。1查询和2查询都比较简单，只是第三步要考虑到时间效率问题。其实这道题还有一个体会就是要领会题目的意思，座位号是101到999的，所以一共只有899个座位，为什么这里要强调，因为第三个查询是要找出该日期下的学号，并且按照座位号进行统计，座位号相同的进行计数，按照数目个数从大到小输出，数目相同，就按照座位号的升序输出。因为座位号只有899个，我在对每一个学号匹配好日期之后，如果是属于该日期下，那么我处理一下座位号，即在相应的座位号进行加一，到时候对座位数组进行排序，时间复杂度就算是899的平方也比10000的平方要小很多。

#include <iostream>
#include<bits/stdc++.h>
#define MaxSize 900
using namespace std;
struct Student
{
    string number;
    int score;
    Student(string temp_number,int temp_score):number(temp_number),score(temp_score) {}
};
struct Site{    
   int site;
   int c_count=0;
};
int N,M;
vector<Student> studentList[3];    //每一个级别单独存放
Site siteList[MaxSize];        //座位号数组，存放该日期下，该座位号有多少人，结构体的site存放是为了排序的时候便于处理
bool Compare(Student s1,Student s2)
{
    if(s1.score>s2.score||(s1.score==s2.score&&(s1.number.compare(s2.number)<0)))
    {
        return true;
    }
    return false;
}
bool Compare_two(Site s1,Site s2){
    if(s1.c_count>s2.c_count||(s1.c_count==s2.c_count&&s1.site<s2.site)){
        return true;
    }
    return false;
}
int Level(char c)
{
    switch(c)
    {
    case 'T':
        return 0;
    case 'A':
        return 1;
    default:
        return 2;
    }
}
void Input()
{
    string tempString;
    int temp_score;
    cin>>N>>M;
    for(int i=0; i<N; i++)
    {
        cin>>tempString>>temp_score;
        switch(tempString[0])
        {
        case 'T':
            studentList[0].push_back(Student(tempString,temp_score));
            break;
        case 'A':
            studentList[1].push_back(Student(tempString,temp_score));
            break;
        default:
            studentList[2].push_back(Student(tempString,temp_score));
        }

    }
}
void Process_One(char level)      //处理1查询
{
    int index=Level(level);
    sort(studentList[index].begin(),studentList[index].end(),Compare);
    if(studentList[index].size()==0){
        cout<<"NA"<<endl;
        return;
    }
    for(int i=0; i<studentList[index].size(); i++)
    {
        cout<<studentList[index][i].number<<" "<<studentList[index][i].score<<endl;
    }
}
void Process_Two(string site)    //处理2查询
{
    int Nt=0,Ns=0;
    int index,k;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<studentList[i].size(); j++)
        {
            index=0;
            for(k=1; k<=3; k++)
            {
                if(studentList[i][j].number[k]!=site[index])
                {
                    break;
                }
                index++;
            }
            if(k==4)
            {
                Nt++;
                Ns+=studentList[i][j].score;
            }
        }
    }
    if(Nt==0){
        cout<<"NA"<<endl;
    }
    else{
        cout<<Nt<<" "<<Ns<<endl;
    }
}

void Process_three(string date){    //处理3查询
    string str;
    int temp_site;
    for(int i=1;i<900;i++){
        siteList[i].site=0;
        siteList[i].c_count=0;
    }
    for(int i=0;i<3;i++){
        for(int j=0;j<studentList[i].size();j++){
           //判断此时的相应位是否日期一致
            str=studentList[i][j].number.substr(4,9);
            str=str.substr(0,6);
            if(str.compare(date)==0){
                temp_site=(studentList[i][j].number[1]-'0')*100+(studentList[i][j].number[2]-'0')*10+studentList[i][j].number[3]-'0';
                siteList[temp_site-100].site=temp_site;
                siteList[temp_site-100].c_count++;
            }
        }
    }
    sort(siteList+1,siteList+900,Compare_two);
    if(siteList[1].c_count==0){
        cout<<"NA"<<endl;
        return;
    }
    int i=1;
    while(siteList[i].c_count>0){
        cout<<siteList[i].site<<" "<<siteList[i].c_count<<endl;
        i++;
    }
}
int main()
{
    int type;
    string diagram;
    Input();
    for(int i=1; i<=M; i++)
    {
        cin>>type>>diagram;
        cout<<"Case "<<i<<": "<<type<<" "<<diagram<<endl;
        switch(type){
         case 1:Process_One(diagram[0]);break;
         case 2:Process_Two(diagram);break;
         default:Process_three(diagram);
        }
    }
    return 0;
}