Updating a Dictionary UVA

In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given an old dictionary and a new dictionary, find out what were changed.

Each dictionary is formatting as follows:

{key:value,key:value,...,key:value}

Each key is a string of lower-case letters, and each value is a non-negative integer without leading zeros or prefix ‘+’. (i.e. -4, 03 and +77 are illegal). Each key will appear at most once, but keys can appear in any order.

Input

The first line contains the number of test cases T (T ≤ 1000). Each test case contains two lines. The first line contains the old dictionary, and the second line contains the new dictionary. Each line will contain at most 100 characters and will not contain any whitespace characters. Both dictionaries could be empty.

WARNING: there are no restrictions on the lengths of each key and value in the dictionary. That means keys could be really long and values could be really large.

Output

For each test case, print the changes, formatted as follows:

• First, if there are any new keys, print ‘+’ and then the new keys in increasing order (lexicographically), separated by commas.

• Second, if there are any removed keys, print ‘-’ and then the removed keys in increasing order (lexicographically), separated by commas.

• Last, if there are any keys with changed value, print ‘*’ and then these keys in increasing order (lexicographically), separated by commas.

If the two dictionaries are identical, print ‘No changes’ (without quotes) instead. Print a blank line after each test case.

Sample Input

3

{a:3,b:4,c:10,f:6}

{a:3,c:5,d:10,ee:4}

{x:1,xyz:123456789123456789123456789}

{xyz:123456789123456789123456789,x:1}

{first:1,second:2,third:3}

{third:3,second:2}

Sample Output

3

{a:3,b:4,c:10,f:6}

{a:3,c:5,d:10,ee:4}

{x:1,xyz:123456789123456789123456789}

{xyz:123456789123456789123456789,x:1}

{first:1,second:2,third:3}

{third:3,second:2}

HINT

思路很清晰，使用map<string,string>来存储键和值，一个用于存储新词典，一个用于储存旧词典。用两个数组分别储存修改的和新增的词。遍历新词典：

如果新旧词典的键和值相同就删除旧词典的词

如果新词典中有而旧词典中也有点值不相同，说明是修改的部分，将词存入数组，在旧词典中删除该词。

如果旧词典中没有该词，说明是新增加的词，存入另一个数组，在旧词典中删除该词。

遍历完成之后，旧词典中剩下的就是新词典中删除的词。

最后输出结果就可以了。

注意：每组数据之后都要输出一个空行，包括最后一行。

Accepted

#include<bits/stdc++.h>

using namespace std;

int main() {

    ofstream fcout;

    fcout.open("temp.txt");

    int num;

    string s, s1, s2;

    map<string, string>old, news;

    vector<string>newkey,rekey;

    cin >> num;getchar();

    while (num--){

        int flag = 0;

        old.clear();news.clear();rekey.clear();newkey.clear();

        cin >> s1 >> s2;

        for (int i = 0;i < s1.length();i++)if (s1[i] == '{' || s1[i] == '}' || s1[i] == ',' || s1[i] == ':')s1[i] = ' ';

        for (int i = 0;i < s2.length();i++)if (s2[i] == '{' || s2[i] == '}' || s2[i] == ',' || s2[i] == ':')s2[i] = ' ';

        stringstream ss1(s1), ss2(s2);

        while (ss1 >> s1 >> s2)old[s1] = s2;

        while (ss2 >> s1 >> s2)news[s1] = s2;

        for (auto i = news.begin();i != news.end();i++ ) {

            if (old.count(i->first) && old[i->first] == i->second)old.erase(i->first);

            else if (old.count(i->first) && old[i->first] != i->second) { rekey.push_back(i->first);old.erase(i->first); }

            else if (!old.count(i->first))newkey.push_back(i->first);

        }

        if (newkey.size()) {

            for (int i = 0;i < newkey.size();i++) cout << (i == 0 ? '+' : ',') << newkey[i];

            cout << endl;

        }

        if (old.size()) {

            for (auto i = old.begin();i != old.end();i++)cout << (i == old.begin() ? '-' : ',') << i->first;

            cout << endl;

        }

        if (rekey.size()) {

            for (int i = 0;i < rekey.size();i++) cout << (i == 0 ? '*' : ',') << rekey[i];

            cout << endl;

        }

        if (!newkey.size() && !rekey.size() && !old.size())cout << "No changes" << endl;

        cout << endl;

    }

}

巴特西

Updating a Dictionary UVA - 12504