Following Orders(拓撲排序)


Description

Order is an important concept in mathematics and in computer science. For example, Zorn's Lemma states: ``a partially ordered set in which every chain has an upper bound contains a maximal element.'' Order is also important in reasoning about the fix-point semantics of programs.  


This problem involves neither Zorn's Lemma nor fix-point semantics, but does involve order.  
Given a list of variable constraints of the form x < y, you are to write a program that prints all orderings of the variables that are consistent with the constraints.  


For example, given the constraints x < y and x < z there are two orderings of the variables x, y, and z that are consistent with these constraints: x y z and x z y.  

Input

The input consists of a sequence of constraint specifications. A specification consists of two lines: a list of variables on one line followed by a list of contraints on the next line. A constraint is given by a pair of variables, where x y indicates that x < y.  


All variables are single character, lower-case letters. There will be at least two variables, and no more than 20 variables in a specification. There will be at least one constraint, and no more than 50 constraints in a specification. There will be at least one, and no more than 300 orderings consistent with the contraints in a specification.  


Input is terminated by end-of-file.  

Output

For each constraint specification, all orderings consistent with the constraints should be printed. Orderings are printed in lexicographical (alphabetical) order, one per line.  


Output for different constraint specifications is separated by a blank line.  

Sample Input

a b f g
a b b f
v w x y z
v y x v z v w v

Sample Output

abfg
abgf
agbf
gabf

wxzvy
wzxvy
xwzvy
xzwvy
zwxvy
zxwvy

 

 

 

 

解題思路:一個一個的添,當填滿的時候輸出,注意添的時候,新添的一個必須沒被用過,並且前面沒比他小的.     比如第二行輸入時a b;代表a在b前面,則填入a時,前面不能有b;


 

#include <iostream>
#include <string>
using namespace std;
const int maxn=30; 
const int maxm=500; 
int n;
int G[maxn][maxn]; 
int vis[maxn]; 
int ans[maxn]; 
int cnt; 
bool mark[maxn];

bool ok(int i,int cnt)
{int j;
 for(j=0;j<cnt;j++)
if(G[i][ans[j]])return 0;
return 1;


}

 

 

void dfs(int cnt)
{int i,j;
 if(cnt==n)
{for(j=0;j<n;j++)
cout<<char(ans[j]+'a');
cout<<endl;}
else{
for(i=0;i<26;i++)
if(mark[i]&&!vis[i]&&ok(i,cnt))
{vis[i]=1;
ans[cnt]=i;
dfs(cnt+1);
vis[i]=0;
}
}
}
int main()
{
char str[1000];
while(gets(str))//注意輸入時有空格;

{int i;

 n=0; 
 memset(mark,0,sizeof(mark)); 
 memset(G,0,sizeof(G)); 
 memset(vis,0,sizeof(vis));
for(i=0;i<str[i];i++)if(str[i]!=' ') 
{mark[str[i]-'a']=true;
n++;}
gets(str);
for( i=0;str[i];i++)if(str[i]!=' ') 
        { 
            int a,b; 
            a=str[i++]-'a'; 
            while(str[i]==' ') 
                i++; 
            b=str[i]-'a'; 
            G[a][b]=1; 
        } 


dfs(0);

cout<<endl;//注意每次輸出后都有一空行
}
return 0;}


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