HDU——2723Electronic Document Security（STL map嵌套set做法）

Electronic Document Security

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 172    Accepted Submission(s): 94

Problem Description

 

The Tyrell corporation uses a state-of-the-art electronic document system that controls all aspects of document creation, viewing, editing, and distribution. Document security is handled via access control lists (ACLs). An ACL defines a set of entities that have access to the document, and for each entity defines the set of rights that it has. Entities are denoted by uppercase letters; an entity might be a single individual or an entire division. Rights are denoted by lowercase letters; examples of rights are a for append, d for delete, e for edit, and r for read.

The ACL for a document is stored along with that document, but there is also a separate ACL log stored on a separate log server. All documents start with an empty ACL, which grants no rights to anyone. Every time the ACL for a document is changed, a new entry is written to the log. An entry is of the form ExR, where E is a nonempty set of entities, R is a nonempty set of rights, and x is either "+", "–", or "=". Entry E+R says to grant all the rights in R to all the entities in E, entry E–R says to remove all the rights in R from all the entities in E, and entry E=R says that all the entities in E have exactly the rights in R and no others. An entry might be redundant in the sense that it grants an entity a right it already has and/or denies an entity a right that it doesn't have. A log is simply a list of entries separated by commas, ordered chronologically from oldest to most recent. Entries are cumulative, with newer entries taking precedence over older entries if there is a conflict.

Periodically the Tyrell corporation will run a security check by using the logs to compute the current ACL for each document and then comparing it with the ACL actually stored with the document. A mismatch indicates a security breach. Your job is to write a program that, given an ACL log, computes the current ACL.

 

 

Input

 

The input consists of one or more ACL logs, each 3–79 characters long and on a line by itself, followed by a line containing only "#" that signals the end of the input. Logs will be in the format defined above and will not contain any whitespace.

 

 

Output

 

For each log, output a single line containing the log number (logs are numbered sequentially starting with one), then a colon, then the current ACL in the format shown below. Note that (1) spaces do not appear in the output; (2) entities are listed in alphabetical order; (3) the rights for an entity are listed in alphabetical order; (4) entities with no current rights are not listed (even if they appeared in a log entry), so it's possible that an ACL will be empty; and (5) if two or more consecutive entities have exactly the same rights, those rights are only output once, after the list of entities.

 

 

Sample Input

 

MC-p,SC+cYB=rde,B-dq,AYM+eGQ+tju,GH-ju,AQ-z,Q=t,QG-tJBL=fwa,H+wf,LD-fz,BJ-a,P=aw#

 

 

Sample Output

 

1:CSc2:AeBerMeYder3:4:BHJfwLPaw

这题是我写过最长的代码题。也是我个大菜鸟写过最复杂的..对于map和set又熟悉了不少。，题意:每一行输入多个names（集合） x rights（集合），name和rights是字符串，x是操作符+、-、=。

举个例子，GOQ=ab，则产生G、O、G三个人（并先清空当前所有权限），每一个人都赋给a与b的权限，题目所给的写法可以视为对一个name集合的每一个字符进行操作，再举个例子,GOQ+ab，则将G、O、Q三个人均加上a与b的权限（若已有则不变）再举个栗子吧GOQ-ab，则将G、O、Q三个人权限中均减掉a与b（若原来就没有就不用减）

题目输出要求：对于具有相同权限的人按字典序排序再合并输出，若某人无任何权限，则不输出。因为按字典序，则用map，又由于每一个人的权限重复删除、增加问题，则用set。当然题目本身后台数据据说不是很强，我的代码也许是有问题的。

#include<iostream>#include<cstdio>#include<string>#include<map>#include<set>using namespace std;int main(void){    string t,s,name,miaoshu;    char na;    int i,j,k,mid,coma,q,len,p=0;    while (getline(cin,t)&&t!="#")    {        p++;//输入计数器        set<char> right;        map<char,set<char> >list;        map<char,set<char> >::iterator mit,tmit;        set<char>::iterator sit;        map<char,set<char> >::reverse_iterator jmit,kmit;        k=0;        while (t.find(",",k)!=string::npos)//由于结尾无逗号，只能先处理前n-1个短句        {            coma=t.find(",",k);            for (i=k; i<coma; i++)            {                if(t[i]=='-')                {                    for (j=k; j<i; j++)                    {                        for(q=i+1; q<coma; q++)                        {                            if(list[t[j]].count(t[q])!=0)                                list[t[j]].erase(t[q]);                        }                                            }                    break;                                }                else if(t[i]=='+')                {                    for (j=k; j<i; j++)                    {                        for(q=i+1; q<coma; q++)                        {                            if(list[t[j]].count(t[q])==0)                                list[t[j]].insert(t[q]);                        }                                            }                    break;                }                else if(t[i]=='=')                {                    for (j=k; j<i; j++)                    {                        list[t[j]].clear();                        for(q=i+1; q<coma; q++)                        {                            list[t[j]].insert(t[q]);                        }                                            }                    break;                }                                }            k=coma+1;        }        //开始处理最后一个句子        len=t.size();        for (i=k; i<len; i++)        {            if(t[i]=='-')            {                for (j=k; j<i; j++)                {                    for(q=i+1; q<len; q++)                    {                        if(list[t[j]].count(t[q])!=0)                            list[t[j]].erase(t[q]);                    }                                        }                break;                            }            else if(t[i]=='+')            {                for (j=k; j<i; j++)                {                    for(q=i+1; q<len; q++)                    {                        if(list[t[j]].count(t[q])==0)                            list[t[j]].insert(t[q]);                    }                                        }                break;            }            else if(t[i]=='=')            {                for (j=k; j<i; j++)                {                    list[t[j]].clear();                    for(q=i+1; q<len; q++)                    {                        list[t[j]].insert(t[q]);                    }                                        }                break;            }                            }            //处理完毕                      for (mit=list.begin(); mit!=list.end(); mit++)//找出没有权限的人，抹掉他        {            if((mit->second).empty())                list.erase(mit++);        }        printf("%d:",p);                for (mit=list.begin(); mit!=list.end(); mit++)        {            tmit=mit;            tmit++;                if((tmit->second)!=(mit->second)&&!(mit->second).empty())//若第n个与第n-1个权限不相同且权限不为空，则说明不用合并            {                cout<<mit->first;                for(sit=(mit->second).begin(); sit!=(mit->second).end(); sit++)                    cout<<*sit;            }            else if(!(mit->second).empty())//若权限相同且不为空则只要输出名字即可                    cout<<mit->first;                }        printf("\n");    }    return 0;}