hdu 2723 Electronic Document Security

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
题意：首先给你文件集合（大写的），然后给你文件的属性（小写字母），‘+’是给文件加上其属性，‘-’是减去其属性，‘=’是把其原来的属性覆盖掉。
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;char s[110],s1[110];int Map[30][30];int i,j,k,l,l1,f;struct node {    char c;    char a[30];} e[30];int main() {    int kase = 0;    while(cin>>s && strcmp(s,"#")){        memset(Map,0,sizeof(Map));//存第i行的j属性是否存在        l=strlen(s);        l1=0;        for(i=0;i<l;i++){            if(s[i]==','){                memset(s1,0,sizeof(s1));                l1=0;            }            else if(s[i]=='+'){                for(j=i+1;  s[j]!=','&& j<l;j++){//找到文件的属性                    for(k=0;k<l1;k++){                        Map[s1[k]-'A'][s[j]-'a']=1;//第i行的j属性存在                    }                }            }            else if(s[i]=='-'){                for(j=i+1;  s[j]!=','&& j<l;j++){                    for(k=0;k<l1;k++){                        Map[s1[k]-'A'][s[j]-'a']=0;//第i行的j属性不存在                    }                }            }            else if(s[i]=='='){                for(k=0;k<l1;k++){                    memset(Map[s1[k]-'A'],0,sizeof(Map[0]));//先把之前存在的属性清空                    for(j=i+1; s[j]!=','&&j<l ;j++)                         Map[s1[k]-'A'][s[j]-'a']=1;                }            }            else s1[l1++]=s[i];//s1存的是文件        }    for(i=0;i<27;i++){          e[i].c=0;          memset(e[i].a,0,sizeof(e[i].a));//赋初始值       }    l=l1=0;    for(i=0; i<27; i++) {        l1=f=0;        for(j=0; j<27; j++) {            if(Map[i][j]>0) {                // printf("i=%d j=%d\n",i,j);                e[l].c=i+'A';                //  printf("e[l].c=%c\n",e[l].c);                e[l].a[l1++]=j+'a';                f=1;            }        }        if(f)            l++;        // if(l1)        // printf("%s*\n",e[l-1].a);    }//把结果从Map里面读出来    printf("%d:",++kase);    for(i=0; i<l; i++) {        printf("%c",e[i].c);        if(strcmp(e[i].a,e[i+1].a)==0)            continue;        else            cout<<e[i].a;    }    printf("\n");    }    return 0;}