Following Orders
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4254 | Accepted: 1709 |
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 收获：1.了解了stringstream. 2.用dfs输出拓扑排序的所有情况。
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#include <sstream>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 500
#define maxm 28
char a[maxn];
char ans[maxn];
int in[maxn];
int vis[maxn];
int map1[maxn][maxn];
int total;
void dfs(int id)
{
if(id == total)
{
ans[id] = ‘\0‘;
puts(ans);
return;
}
for(int i = 0; i < 26; i++)
{
if(vis[i]) continue;
if(in[i] == 0)
{
ans[id] = ‘a‘ + i;
vis[i] = 1;
for(int j = 0; j < 26; j++)
if(map1[i][j]) in[j]--;
dfs(id+1);
vis[i] = 0;
for(int j = 0; j < 26; j++)
if(map1[i][j]) in[j]++;
}
}
}
char x, y;
int main()
{
int flag = 0;
while(gets(a) != NULL)
{
if(flag)
puts("");
flag = 1;
total = 0;
stringstream ss(a);
memset(in, INF, sizeof in);
memset(vis, 0, sizeof vis);
while(ss >> x)
{
in[x - ‘a‘] = 0;
total++;
}
gets(a);
stringstream sss(a);//读取一行。
memset(map1, 0, sizeof map1);
while(sss >> x >> y)//扫描该行的字符。
{
map1[x - ‘a‘][y- ‘a‘] = 1;
in[y - ‘a‘]++;
}
dfs(0);
}
return 0;
}