A1153 Decode Registration Card of PAT（25分）PAT甲级 (Advanced Level) Practice（C++）满分题解【unordered

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 (≤104) 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

这道题最要注意的是超时问题！！！

如果你也是超时的情况，以我出错的地方来看：

（1）将map类型修改为unordered_map

（2）将字符串的输出从cout修改为printf

代码如下：

#include
using namespace std;
struct Card {
	string level, site, date, number;
	int score;
};
bool cmp(Card &a, Card &b)
{
	if (a.score != b.score)return a.score > b.score;
	else {
		string sa = a.level + a.site + a.date + a.number;
		string sb = b.level + b.site + b.date + b.number;
		return sa < sb;
	}
}
bool cmp2(Card &a, Card &b)
{
	if (a.score != b.score)return a.score > b.score;
	else return a.site> levels, sites, dates;
int main()
{
	int n, m, s;
	cin >> n >> m;
	string str;
	for (int i = 0; i < n; i++)
	{
		cin >> str >> s;
		Card temp;
		temp.score = s;
		temp.level = str.substr(0, 1);
		temp.site = str.substr(1, 3);
		temp.date = str.substr(4, 6);
		temp.number = str.substr(10, 3);
		levels[temp.level].push_back(temp);
		sites[temp.site].push_back(temp);
		dates[temp.date].push_back(temp);
	}
	string type, content;
	for (int i = 0; i < m; i++)
	{
        bool flag=false;
		cin>>type>>content;
		cout << "Case "< cnt;
            vector ans;
            if(dates[content].size()!=0)flag=true;
			for (int j = 0; j < dates[content].size(); j++) {//在该日期里找每个网站的个数
				cnt[dates[content][j].site]++;
			}
			for (unordered_map::iterator it = cnt.begin(); it != cnt.end(); it++){
                Card tmp;
                tmp.site=it->first;
                tmp.score=it->second;
                ans.push_back(tmp);
            }
            sort(ans.begin(),ans.end(),cmp2);
            for(int j=0;j

运行结果如下：