POJ1270 Following Orders（拓扑排序）
来源：程序员人生 发布时间：2015-09-07 08:27:00 阅读次数：4188次
Following Orders
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 4059 |
|
Accepted: 1623 |
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
Source
Duke Internet Programming Contest 1993,uva 124
题意：给定1串字符（互异），再给出1个字符序列，表示1种前后关系，如a b e f c d,表示a<b,e<f,c<d。
将开始给出的字符进行排序，使之符合这个关系序列。并按字典序输出这些符合要求的字符序列。
#include<stdio.h>
#include<vector>
#include<string.h>
using namespace std;
const int N = 30;
int in[N],exist[N],mapt[N][N],path[N],n;
void topeSort(int u,int k)
{
path[k]=u;
if(k==n)
{
for(int i=1;i<=n;i++)
printf("%c",path[i]+'a');
printf("
");
return ;
}
in[u]=⑴;
for(int i=0;i<26;i++)
if(mapt[u][i])
in[i]-=mapt[u][i];
for(int i=0;i<26;i++)
if(exist[i]&&!in[i])
topeSort(i,k+1);
in[u]=0;
for(int i=0;i<26;i++)
if(mapt[u][i])
in[i]+=mapt[u][i];
}
int main()
{
int flag=0;
char str[1000];
while(gets(str))
{
if(flag)printf("
"); flag=1;
memset(in,0,sizeof(in));
memset(exist,0,sizeof(exist));
memset(mapt,0,sizeof(mapt));
n=0;
for(int i=0;str[i]!=' ';i++)
if(str[i]>='a'&&str[i]<='z')
{
int ch=str[i]-'a';
if(exist[ch]==0)
n++;
exist[ch]=1;
}
gets(str);
int i=0,a,b;
while(str[i]!=' ')
{
while(str[i]==' ')i++; a=str[i++]-'a';
while(str[i]==' ')i++; b=str[i++]-'a';
mapt[a][b]++; in[b]++;
}
for(int i=0;i<26;i++)
if(exist[i]&&!in[i])
topeSort(i,1);
}
}
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