CSU 1113 Updating a Dictionary

最新推荐文章于 2018-04-22 21:31:04 发布

转载最新推荐文章于 2018-04-22 21:31:04 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/Patt/p/4730103.html

本文解析了一道关于字典对比的算法题，通过输入两个字典格式的字符串，找出新增、删除及变更的键值对，并按指定格式输出。文章提供了错误及正确的代码示例，有助于理解实现细节。

传送门

Time Limit: 1000MS

Memory Limit: 131072KB

64bit IO Format: %lld & %llu

Description

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

+d,ee
-b,f
*c

No changes

-first

------------------------------------------------------------------------------

简单题，注意implementation。

------------------------------------------------------------------------------

#include <cstdio>
#include <iostream>
#include <cctype>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#define pb push_back

using namespace std;
map<string, string> a, b;
vector<string> aa, bb;
int main(){
	//freopen("in", "r", stdin);
	int T;
	string s, t;
	char ch;
	for(cin>>T; T--; cout<<endl){
		for(;cin>>ch;){
			if(isalpha(ch)) s+=ch;
			else if(ch==':'){
				while(cin>>ch, isdigit(ch)){
					t+=ch;
				}
				a[s]=t;
				aa.pb(s);
				s.clear();
				t.clear();
				if(ch=='}') break;	//error
			}
		}
		for(;cin>>ch;){
			if(isalpha(ch)) s+=ch;
			else if(ch==':'){
				while(cin>>ch, isdigit(ch)){
					t+=ch;
				}
				bb.pb(s);
				b[s]=t;
				s.clear();
				t.clear();
				if(ch=='}') break;	//error
			}
		}
		sort(aa.begin(), aa.end());
		sort(bb.begin(), bb.end());
		bool fir=true, nc=true;
		for(int i=0; i<bb.size(); i++){
			string &tmp=bb[i];
			if(a[tmp].empty()){
				if(fir) cout<<'+'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		fir=true;
		for(int i=0; i<aa.size(); i++){
			string &tmp=aa[i];
			if(b[tmp].empty()){
				if(fir) cout<<'-'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		fir=true;
		for(int i=0; i<bb.size(); i++){
			string &tmp=bb[i];
			if(!a[tmp].empty()&&!b[tmp].empty()&&a[tmp]!=b[tmp]){
				if(fir) cout<<'*'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		if(nc) cout<<"No changes"<<endl;
		a.clear(), b.clear(), aa.clear(), bb.clear();
	}

}

#include <cstdio>
#include <iostream>
#include <cctype>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#define pb push_back

using namespace std;
map<string, string> a, b;
vector<string> aa, bb;
int main(){
	freopen("in", "r", stdin);
	int T;
	string s, t;
	char ch;
	for(cin>>T; T--; cout<<endl){
		for(;cin>>ch;){
			if(isalpha(ch)) s+=ch;
			else if(ch==':'){
				while(cin>>ch, isdigit(ch)){
					t+=ch;
				}
				a[s]=t;
				aa.pb(s);
				s.clear();
				t.clear();
			}
			if(ch=='}') break;
		}
		//cout<<aa.size()<<endl;
		for(;cin>>ch;){
			if(isalpha(ch)) s+=ch;
			else if(ch==':'){
				while(cin>>ch, isdigit(ch)){
					t+=ch;
				}
				bb.pb(s);
				b[s]=t;
				s.clear();
				t.clear();
			}
			if(ch=='}') break;
		}
		sort(aa.begin(), aa.end());
		sort(bb.begin(), bb.end());
		bool fir=true, nc=true;
		for(int i=0; i<bb.size(); i++){
			string &tmp=bb[i];
			if(a[tmp].empty()){
				if(fir) cout<<'+'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		fir=true;
		for(int i=0; i<aa.size(); i++){
			string &tmp=aa[i];
			if(b[tmp].empty()){
				if(fir) cout<<'-'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		fir=true;
		for(int i=0; i<bb.size(); i++){
			string &tmp=bb[i];
			if(!a[tmp].empty()&&!b[tmp].empty()&&a[tmp]!=b[tmp]){
				if(fir) cout<<'*'+tmp, fir=false;
				else cout<<','+tmp;
			}
		}
		if(!fir) cout<<endl, nc=0;
		if(nc) cout<<"No changes"<<endl;
		a.clear(), b.clear(), aa.clear(), bb.clear();
	}
}