Following Orders
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4885 | Accepted: 1973 |
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.
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.
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.
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
－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－
所有方案，需要回溯，用Kahn比較好
L← Empty list that will contain the sorted elements S ← Set of all nodes with no incoming edges while S is non-empty do remove a node n from S insert n into L foreach node m with an edge e from nto m do remove edge e from thegraph ifm has no other incoming edges then insert m into S if graph has edges then return error (graph has at least onecycle) else return L (a topologically sortedorder)
就是找入度為0的點(最好用個stack,迴圈的話複雜的太高)，加入topo頭部
感覺比dfs好，複雜度都是O(V+E)
本題回溯所有方案，複雜度乘上一個V；V很小，不用stack也可以；用個id比較方便吧
字元讀入太坑人.........
// // main.cpp // poj1270 // // Created by Candy on 9/11/16. // #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N=30,M=55; char s[105]; int a[N],num=0,n=0,id[N]; int ch[N][N],topo[N],ind[N]; void print(){ for(int i=1;i<=n;i++) printf("%c",(char)topo[i]+'a'-1); printf("\n"); } void dfs(int d){ //printf("dfs %d\n",d); if(d==n+1){print();return;} for(int i=1;i<=n;i++) if(ind[i]==0){ ind[i]--; topo[d]=a[i]; for(int j=1;j<=ch[i][0];j++) ind[ch[i][j]]--; dfs(d+1); for(int j=1;j<=ch[i][0];j++) ind[ch[i][j]]++; ind[i]++; } } int main(int argc, const char * argv[]) { while(fgets(s,105,stdin)){ //printf("p %s\n",s); n=0; memset(topo,0,sizeof(topo)); memset(ch,0,sizeof(ch)); memset(ind,0,sizeof(ind)); int len=strlen(s); //printf("len %d\n",len); for(int i=0;i<len;i++) if(s[i]>='a'&&s[i]<='z') a[++n]=s[i]-'a'+1; sort(a+1,a+1+n); for(int i=1;i<=n;i++) id[a[i]]=i; fgets(s,105,stdin); len=strlen(s); int last=0; for(int i=0;i<=len;i++) if(s[i]>='a'&&s[i]<='z'){ int t=s[i]-'a'+1; t=id[t]; if(last==0) last=t; else{ch[last][++ch[last][0]]=t;ind[t]++;last=0;} } dfs(1); printf("\n"); } return 0; }