POJ 1270 Following Orders 拓扑排序全输出

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：给出一些字母 ， 和字母间关系 ，让你输出所有符合关系的排序
分析：相当于是按字典序输出所有的拓扑排序情况。

鉴于要输出所有情况，那么我在拓扑排序时给每个字母设置一个等级，在用next_permutation找到所有排列，只要复合拓扑排序规则，这个排列一定是等级递增关系，直接输出。
本以为这样做会T，没想到数据太弱，0msAC。

  1 #include<cstring>
  2 #include<cstdio>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<map>
  7 #include<vector>
  8 #include<cstdlib>
  9 
 10 using namespace std;
 11 
 12 char a[105],b[105];
 13 int vis[105];
 14 int in[105];
 15 vector<int>out[105];
 16 vector<int>chat;
 17 vector<int>temp;
 18 
 19 void init()
 20 {
 21     for(int i=0;i<26;i++)
 22     {
 23         in[i]=0;
 24         out[i].clear();
 25         vis[i]=-2;
 26     }
 27     chat.clear();
 28     temp.clear();
 29 }
 30 
 31 void toposort()
 32 {
 33     int num=0;
 34     int cur=0;
 35     for(int i=0;i<chat.size();i++) if(vis[chat[i]]==-2) num++;
 36 
 37     while(num)
 38     {
 39 
 40         for(int i=0;i<chat.size();i++)
 41         {
 42             if(in[chat[i]]==0&&vis[chat[i]]!=-1)
 43             {
 44                 in[chat[i]]--;
 45                 temp.push_back(chat[i]);
 46             }
 47         }
 48 
 49         for(int i=0;i<temp.size();i++)
 50         {
 51             int k=temp[i];
 52             vis[k]=cur;
 53             for(int j=0;j<out[k].size();j++)
 54             {
 55                 in[out[k][j]]--;
 56             }
 57         }
 58 
 59         num-=temp.size();
 60         temp.clear();
 61         cur++;
 62 
 63     }
 64 }
 65 
 66 bool judge()
 67 {
 68     int maxn=0;
 69     for(int i=0;i<chat.size();i++)
 70     {
 71         if(vis[chat[i]]!=-1)
 72         {
 73             if(vis[chat[i]]>=maxn) maxn=vis[chat[i]];
 74             else return false;
 75         }
 76     }
 77     return true;
 78 }
 79 
 80 int main()
 81 {
 82     int black=1;
 83     while(gets(a)!=NULL)
 84     {
 85         gets(b);
 86 
 87         if(black==0) printf("
");
 88         init();
 89 
 90         int len=strlen(a);
 91         for(int i=0;i<len;i++)
 92         {
 93             if(a[i]>='a'&&a[i]<='z')
 94             {
 95                 chat.push_back(a[i]-'a');
 96             }
 97         }
 98 
 99         len=strlen(b);
100         char k='#';
101         for(int i=0;i<len;i++)
102         {
103             if(b[i]>='a'&&b[i]<='z')
104             {
105                 if(k=='#') k=b[i];
106                 else
107                 {
108                     in[b[i]-'a']++;
109                     out[k-'a'].push_back(b[i]-'a');
110                     k='#';
111                 }
112             }
113         }
114 
115         for(int i=0;i<chat.size();i++)
116             if(!in[chat[i]]&&!out[chat[i]].size()) vis[chat[i]]=-1;
117 
118         toposort();
119 
120         sort(chat.begin(),chat.end());
121         do
122         {
123             if(judge())
124             {
125                 for(int i=0;i<chat.size();i++)
126                     printf("%c",chat[i]+'a');
127                 puts("");
128             }
129         }while(next_permutation(chat.begin(),chat.end()));
130 
131         black=0;
132     }
133 }