[刷题]算法竞赛入门经典(第2版) 5-11/UVa12504 - Updating a Dictionary

题意：对比新老字典的区别：内容多了、少了还是修改了。

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代码：（Accepted，0.000s）

//UVa12504 - Updating a Dictionary
//#define _XieNaoban_
#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<map>
using namespace std;

int T;
int main()
{
#ifdef _XieNaoban_
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin >> T;
    cin.ignore();
    while (T--) {
        map<string, string> dic;
        for (int i(0);i < 2;++i) {
            string line, key, value;
            getline(cin, line);
            for (auto &r : line)
                if (r == ',' || r == ':' || r == '{' || r == '}') r = ' ';
            istringstream in(line);
            while (in >> key) {
                in >> value;
                if (i) {
                    auto d(dic.find(key));
                    if (d != dic.end()) {
                        if (d->second != value) d->second = "*";//changed
                        else d->second = '&';//not changed
                    }
                    else dic[key] = "+";//increased
                }
                else dic[key] = value;
            }
        }
        vector<string> inc, rmv, chg;
        for (const auto& r : dic)
            switch (r.second[0])
            {
            case '+': inc.push_back(r.first);break;
            case '*': chg.push_back(r.first);break;
            case '&': break;
            default:  rmv.push_back(r.first);break;
            }
        if (inc.empty() && rmv.empty() && chg.empty())
            cout << "No changes\n";
        else {
            if (!inc.empty()) {
                cout << '+' << inc[0];
                for (auto i(inc.begin() + 1);i != inc.end();++i)
                    cout << ',' << *i;
                cout << '\n';
            }
            if (!rmv.empty()) {
                cout << '-' << rmv[0];
                for (auto i(rmv.begin() + 1);i != rmv.end();++i)
                    cout << ',' << *i;
                cout << '\n';
            }
            if (!chg.empty()) {
                cout << '*' << chg[0];
                for (auto i(chg.begin() + 1);i != chg.end();++i)
                    cout << ',' << *i;
                cout << '\n';
            }
        }
        cout << '\n';
    }
    return 0;
}

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分析：这题很水，轻松AC，没啥好说的。