poj1207 拓扑排序+dfs

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

 题意L:输入数据有两行,第一行给你由26个单个小写字母表示的变量,第二行给出成对的变量如(x,y),表示x要在y前面.然后要你输出所有可能的变量序列且满足第二行的顺序约束.如果存在多解,按字典序从小到大输出所有结果.

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=30;
const int maxm=500;
int n;//有效字母数
int G[maxn][maxn];
int vis[maxn];
int ans[maxn];
int cnt;
bool mark[maxn];//标记当前字母出现在变量中
bool ok(int i,int cnt)//如果在ans[0,cnt-1]出现了一个本应在i后面才出现的字母,那么返回false
{
    for(int j=0;j<cnt;j++)
        if(G[i][ans[j]]) return false;
    return true;
}
void dfs(int cnt)
{
    if(cnt==n)
    {
        for(int i=0;i<n;i++)
            printf("%c",ans[i]+'a');
        printf("\n");
    }
    else for(int 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))
    {
        n=0;
        memset(mark,0,sizeof(mark));
        memset(G,0,sizeof(G));
        memset(vis,0,sizeof(vis));
        for(int i=0;str[i];i++)if(str[i]!=' ')
            mark[str[i]-'a']=true, n++;
        gets(str);
        for(int 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);//表示当前正在构造第0个位置
        puts("");
    }
    return 0;
}