题意  比较两个字典  按字典序输出所有添加 删除 修改的项   如果没有任何更新  输出  No changes

STL map的应用  对比两个字典  注意开始字符串的处理和字典可以为空

#include<bits/stdc++.h>
using namespace std;
map<string, string> d[2];
map<string, string>::iterator it;
const int N = 105;
string s, a, b, t[N];

void print(char c, int n)
{
    sort(t, t + n), cout << c << t[0];
    for(int i = 1; i < n; ++i) cout << ',' << t[i];
    puts("");
}

int main()
{
    int cas, n, c1, c2, c3;
    cin >> cas;
    while(cas--)
    {
        d[0].clear(), d[1].clear();
        for(int i = 0; i < 2; ++i)
        {
            cin >> s;
            int j = 1, l = s.size();
            while(l > 2 && j < l)
            {
                while(s[j] != ':') a += s[j++]; ++j;
                while(s[j] != ',' && s[j] != '}') b += s[j++]; ++j;
                d[i][a] = b, a = b = "";
                //cout << a << " : " << b << endl;
            }
        }

        c1 = c2 = c3 = 0;
        for(it = d[1].begin(); it != d[1].end(); ++it)
            if(!d[0].count(it->first)) t[c1++] = it->first;
        if(c1) print('+', c1);

        for(it = d[0].begin(); it != d[0].end(); ++it)
            if(!d[1].count(it->first)) t[c2++] = it->first;
        if(c2) print('-', c2);

        for(it = d[1].begin(); it != d[1].end(); ++it)
            if(d[0].count(it->first) && d[0][it->first] != it->second) t[c3++] = it->first;
        if(c3) print('*', c3);

        if(!(c1 || c2 || c3)) puts("No changes");
        puts("");
    }
    return 0;
}

Updating a Dictionary

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:

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 

T1000

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 

• 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 

+d,ee
-b,f
*c

No changes

-first

UVa 12504 Updating a Dictionary(更新字典)