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
思路：
简单的拓扑排序+dfs，通过入度判断，从1到26就有字典序，代码如下：
int in[30], G[30][30], vis[30], print[30], num;
char ans[25];
void init() {
num = 0;
memset(in, 0, sizeof(in));
memset(G, 0, sizeof(G));
memset(vis, 0, sizeof(vis));
memset(print, 0, sizeof(print));
}
void dfs(int u, int cnt) {
ans[cnt] = u - 1 + 'a';
if(cnt == num) {
for(int i = 1; i <= cnt; ++i)
cout << ans[i];
cout << "\n";
return;
}
// mark the point
print[u] = 1;
for(int i = 1; i <= 26; ++i) {
if(vis[i] && G[u][i]) in[i]--;
}
for(int i = 1; i <= 26; ++i) {
if(vis[i] && !print[i] && !in[i]) {
dfs(i, cnt+1);
}
// backtracing
}
for(int i = 1; i <= 26; ++i) {
if(vis[i] && G[u][i]) in[i]++;
}
print[u] = 0;
}
int main() {
ios::sync_with_stdio(false);
string t;
int t1, t2;
while(getline(cin, t)) {
init();
int siz = t.size();
for(int i = 0; i < siz; i += 2) {
vis[t[i] - 'a' + 1] = 1;
num++;
}
getline(cin, t);
siz = t.size();
for(int i = 0; i + 2 < siz; i += 4) {
t1 = t[i] - 'a' + 1, t2 = t[i+2] - 'a' + 1;
G[t1][t2] = 1, in[t2]++;
}
for(int i = 1; i <= 26; ++i) {
if(vis[i] && !in[i]) {
dfs(i, 1);
}
}
cout << "\n";
}
return 0;
}
View Code