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。 

------------------------------------------------------------------------------
 WA
#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=; 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=;
fir=true;
for(int i=; 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=;
fir=true;
for(int i=; 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=;
if(nc) cout<<"No changes"<<endl;
a.clear(), b.clear(), aa.clear(), bb.clear();
} }
 AC 
#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=; 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=;
fir=true;
for(int i=; 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=;
fir=true;
for(int i=; 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=;
if(nc) cout<<"No changes"<<endl;
a.clear(), b.clear(), aa.clear(), bb.clear();
}
}

CSU 1113 Updating a Dictionary的更多相关文章

  1. CSU 1113 Updating a Dictionary(map容器应用)

    题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1113 解题报告:输入两个字符串,第一个是原来的字典,第二个是新字典,字典中的元素的格式为 ...

  2. csuoj 1113: Updating a Dictionary

    http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1113 1113: Updating a Dictionary Time Limit: 1 Sec  ...

  3. 湖南生第八届大学生程序设计大赛原题 C-Updating a Dictionary(UVA12504 - Updating a Dictionary)

    UVA12504 - Updating a Dictionary 给出两个字符串,以相同的格式表示原字典和更新后的字典.要求找出新字典和旧字典的不同,以规定的格式输出. 算法操作: (1)处理旧字典, ...

  4. [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. 代码:(Accepted,0.000s) //UVa12504 - Updating a Dictionary //#define _XieNao ...

  5. [ACM_模拟] UVA 12504 Updating a Dictionary [字符串处理 字典增加、减少、改变问题]

      Updating a Dictionary  In this problem, a dictionary is collection of key-value pairs, where keys ...

  6. Problem C Updating a Dictionary

    Problem C     Updating a Dictionary In this problem, a dictionary is collection of key-value pairs, ...

  7. Updating a Dictionary UVA - 12504

    In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, a ...

  8. Uva 511 Updating a Dictionary

    大致题意:用{ key:value, key:value, key:value }的形式表示一个字典key表示建,在一个字典内没有重复,value则可能重复 题目输入两个字典,如{a:3,b:4,c: ...

  9. Uva - 12504 - Updating a Dictionary

    全是字符串相关处理,截取长度等相关操作的练习 AC代码: #include <iostream> #include <cstdio> #include <cstdlib& ...

随机推荐

  1. ArcGIS Engine 中 Geometric Network 显示流向代码

    原文地址:http://hi.baidu.com/steeeeps/item/165fbc15475e94741009b5b3 非常感谢作者. 以前学习几何网络时,对效用网络流向进行了总结,原理与效果 ...

  2. TIF、JPG图片手动添加地理坐标的方法(转载)

    题目:为TIF.JPG图片添加地理坐标/平面直角坐标. 图片来源:GOOGLE EARTH.(当然也可以是其他知道四角点坐标的图片) 截图工具:GEtscreen(此软件截图时可以自动生成图片四角点坐 ...

  3. Jython概要

    1.安装jython 1.1 进入http://www.jython.org/downloads.html ,网页上会显示当前最稳定的版本(The most current stable releas ...

  4. C语言 位运算

    1G=1024M; 1M=102KB; 1KB=1024B(字节); 1B=8bits(位); #include<stdio.h> #include<stdlib.h> //C ...

  5. 解决SQL Server 阻止了对组件 'Ad Hoc Distributed Queries' 的 STATEMENT'OpenRowset/OpenDatasource' 的访问的方法

    1.开启Ad Hoc Distributed Queries组件,在sql查询编辑器中执行如下语句: reconfigure reconfigure 2.关闭Ad Hoc Distributed Qu ...

  6. python package 的两种组织方式

    方式一/package1/ .../__init__.py # 空文件 .../class1.py class Class1: def __init__(self): self.name = &quo ...

  7. Google java代码风格导入Eclipse

    Git地址 https://github.com/codeset/google-java-styleguide 下载配置文件在Eclipse中执行导入:Window -> Preferences ...

  8. java实现八皇后问题(递归和循环两种方式)

    循环方式: package EightQueens;   public class EightQueensNotRecursive { private static final boolean AVA ...

  9. 利用JS跨域做一个简单的页面访问统计系统

    其实在大部分互联网web产品中,我们通常会用百度统计或者谷歌统计分析系统,通过在程序中引入特定的JS脚本,然后便可以在这些统计系统中看到自己网站页面具体的访问情况.但是有些时候,由于一些特殊情况,我们 ...

  10. C#基础之枚举

    1.认识Enum 以前一直以为Enum是值类型,在VS中查看Enum的定义时才发现它是一个抽象的类.但是这个类很奇怪,Enum继承了ValueType这个很熟悉的值类型基类,它是唯一一个继承自Valu ...