湖南省第八届程序设计竞赛C

Problem C. Updating a Dictionary

 

In this problem, a dictionary is collectionof key-value pairs, where keys are lower-case letters, and values are non-negativeintegers. Given an old dictionary and a new dictionary, find out what werechanged.

 

Each dictionary is formatting asfollows:

 

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

 

Each key is a string oflower-case letters, and each value is a non-negative integer without leadingzeros or prefix '+'. (i.e. -4, 03 and +77 are illegal). Each key will appear atmost once, but keys can appear in any order.

 

Input

The first line contains thenumber of test cases T (T<=1000). Each test case contains twolines. The first line contains the old dictionary, and the second line containsthe new dictionary. Each line will contain at most 100 characters and will notcontain any whitespace characters. Both dictionaries could be empty.

 

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

 

Output

For eachtest case, print the changes, formatted as follows:

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

l  Second, if there are any removedkeys, print '-' and then the removed keys in increasing order(lexicographically), separated by commas.

l  Last, if there are any keys withchanged value, print '*' and then these keys in increasing order(lexicographically), separated by commas.

 

If thetwo dictionaries are identical, print 'No changes' (without quotes) instead.

 

Print ablank line after each test case.

 

Sample Input                                       Output for 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}

+d,ee -b,f *c   No changes   -first

代码：

// Rujia Liu
#include<iostream>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;

void make_dict(string s, map<string,string>& m, vector<string>& keys) {
  for(int i = 0; i < s.size(); i++) {
    if(!isalpha(s[i]) && !isdigit(s[i])) s[i] = ' ';
  }
  stringstream ss(s);
  string key, value;
  while(ss >> key) {
    ss >> value;
    m[key] = value;
    keys.push_back(key);
  }
}

void print_list(const vector<string> v) {
  for(int i = 0; i < v.size(); i++) {
    if(i != 0) cout << ',';
    cout << v[i];
  }
  cout << '\n';
}

int main() {
  int T;
  cin >> T;
  while(T--) {
    string d1, d2;
    map<string,string> m1, m2;
    vector<string> all, added, removed, changed;
    cin >> d1 >> d2;
    make_dict(d1, m1, all);
    make_dict(d2, m2, all);
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());
    for(int i = 0; i < all.size(); i++) {
      string key = all[i];
      if(!m1.count(key) && m2.count(key)) added.push_back(key);
      if(m1.count(key) && !m2.count(key)) removed.push_back(key);
      if(m1.count(key) && m2.count(key) && m1[key] != m2[key]) changed.push_back(key);
    }
    int flag = 0;
    if(!added.empty()) { flag = 1; cout << "+"; print_list(added); }
    if(!removed.empty()) { flag = 1; cout << "-"; print_list(removed); }
    if(!changed.empty()) { flag = 1; cout << "*"; print_list(changed); }
    if(!flag) cout << "No changes\n";
    cout << "\n";
  }
  return 0;
}