Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem A: Open Source

At an open-source fair held at a major university, leaders of open-source projects put sign-up sheets on the wall, with the project name at the top in capital letters for identification.

Students then signed up for projects using their userids. A userid is a string of lower-case letters and numbers starting with a letter.

The organizer then took all the sheets off the wall and typed in the information.

Your job is to summarize the number of students who have signed up for each project. Some students were overly enthusiastic and put their name down several times for the same project. That's okay, but they should only count once. Students were asked to commit to a single project, so any student who has signed up for more than one project should not count for any project.

There are at most 10,000 students at the university, and at most 100 projects were advertised.

Input Specification:

The input contains several test cases, each one ending with a line that starts with the digit 1. The last test case is followed by a line starting with the digit 0.

Each test case consists of one or more project sheets. A project sheet consists of a line containing the project name in capital letters, followed by the userids of students, one per line.

Output Specification:

For each test case, output a summary of each project sheet. The summary is one line with the name of the project followed by the number of students who signed up. These lines should be printed in decreasing order of number of signups. If two or more projects have the same number of signups, they should be listed in alphabetical order.

Sample input

UBQTS TXT
tthumb
LIVESPACE BLOGJAM
philton
aeinstein
YOUBOOK
j97lee
sswxyzy
j97lee
aeinstein
SKINUX
1
0

Output for Sample Input

YOUBOOK 2
LIVESPACE BLOGJAM 1
UBQTS TXT 1
SKINUX 0

---

題目不難理解, 但要做起來又十分麻煩, 而且輸出又要排序,
真是把 STL 一次複習了好多東西

#include <iostream>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
struct D {
    string s;
    int n;
};
bool cmp(D i, D j) {
    if(i.n != j.n)
        return i.n > j.n;
    return i.s.compare(j.s) < 0;
}
int main() {
    while(1) {
        string proj, name;
        map<string, string> stu;
        map<string, set<string> > pro;
        getline(cin, proj);
        stu[""] = "-1";
        while(1) {
            if(proj[0] == '0')
                return 0;
            if(proj[0] == '1')
                break;
            pro[proj].insert("");
            while(getline(cin, name)) {
                if(name[0] >= 'A' && name[0] <= 'Z' || name[0] == '1') {
                    proj = name;
                    break;
                }
                if(stu[name] == "" || stu[name] == proj) {
                    stu[name] = proj;
                    pro[proj].insert(name);
                } else {
                    stu[name] = "-1";
                }
            }
        }
        int idx = 0;
        D ans[100];
        for(map<string, set<string> >::iterator i = pro.begin(); i != pro.end(); i++) {
            //cout << i->first << endl;
            int cnt = 0;
            for(set<string>::iterator j = i->second.begin(); j != i->second.end(); j++) {
                //cout << *j << endl;
                if(stu[*j] != "-1")
                    cnt++;
            }
            ans[idx].s = i->first;
            ans[idx].n = cnt;
            idx++;
        }
        sort(ans, ans+idx, cmp);
        for(int i = 0; i < idx; i++)
            cout << ans[i].s << " " << ans[i].n << endl;
    }
    return 0;
}