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POJ 3654 & ZOJ 2936 & HDU 2723 Electronic Document Security（模拟）
题目链接：
PKU:http://poj.org/problem?id=3654
ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2936
HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2723
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+c YB=rde,B-dq,AYM+e GQ+tju,GH-ju,AQ-z,Q=t,QG-t JBL=fwa,H+wf,LD-fz,BJ-a,P=aw #
Sample Output
1:CSc 2:AeBerMeYder 3: 4:BHJfwLPaw
Source
题意：
运算符（“‘+’， ‘-’, ‘‘=”）前面是名字，‘+’表示把后面的权力赋给前面的人，‘-’表示从前面的人中减去后面的权力（当然没有就不减），‘=’表示把前面的人的权力先清空在把后面的权力赋给前面的人！输出的时候有相同权力的相邻两个人就名字一起输出，再一起输出权力！如果都没有权力则不输出！
代码如下：
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
using namespace std;
const int maxn = 57;
int a[maxn][maxn];
int main()
{
string tt;
char str[1017];
int cas = 0;
int num[maxn];
while(scanf("%s",str)!=EOF)
{
int len = strlen(str);
if(len == 1 && str[0] == '#')
break;
tt = "";
memset(a,0,sizeof a);
for(int i = 0; i < len; i++)
{
if(str[i]==',')
tt = "";//清空
else if(str[i] == '+')
{
for(int j = i+1; str[j]!=',' && j < len; j++)
{
for(int k = 0; k < tt.size(); k++)
{
a[tt[k]-'A'][str[j]-'a']=1;
}
}
}
else if(str[i] == '-')
{
for(int j = i+1; str[j]!=',' && j < len; j++)
{
for(int k = 0; k < tt.size(); k++)
{
a[tt[k]-'A'][str[j]-'a']=0;
}
}
}
else if(str[i] == '=')
{
for(int k = 0; k < tt.size(); k++)
{
memset(a[tt[k]-'A'],0,sizeof a[0]);
for(int j = i+1; str[j]!=',' && j < len; j++)
{
a[tt[k]-'A'][str[j]-'a']=1;
}
}
}
else
tt+=str[i];
}
printf("%d:",++cas);
int l = 0;
for(int i = 0; i < 26; i++)//记录下所有有权力的人
{
int flag = 0;
for(int j = 0; j < 26; j++)
{
if(a[i][j])
flag = 1;
}
if(flag)
{
num[l++] = i;
}
}
tt = "";
for(int i = 0; i < l; i++)
{
int flag = 0;
if(i != l-1)
{
for(int j = 0; j < 26; j++)
{
if(a[num[i]][j] != a[num[i+1]][j])//所含的权力不同
{
flag = 1;
}
}
}
char tem = 'A'+num[i];
tt += tem;
if(flag==1 || i==l-1)
{
cout<<tt;
tt = "";
for(int j = 0; j < 26; j++)
{
if(a[num[i]][j])
printf("%c",'a'+j);
}
}
}
printf("\n");
}
return 0;
}
POJ 3654 & ZOJ 2936 & HDU 2723 Electronic Document Security（模拟）