题目传送门
1 /*
2 题意：将字符串分割成一个全排列
3 DFS：搜索主要在一位数和两位数的处理，用d1, d2记录个数，在不饱和的情况下，两种都试一下
4 DFS还是写不来，难道是在家里懒？
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cstring>
9 #include <cmath>
10 using namespace std;
11
12 const int MAXN = 1e2 + 10;
13 const int INF = 0x3f3f3f3f;
14 char s[MAXN];
15 bool vis[55];
16 int ans[55];
17 int len, tot, n1, n2;
18
19 bool DFS(int d1, int d2, int p, int cnt)
20 {
21 if (cnt == tot) return true;
22 if (d1 < n1)
23 {
24 int x = s[p] - '0';
25 if (x != 0 && !vis[x])
26 {
27 vis[x] = true; ans[cnt+1] = x;
28 if (DFS (d1+1, d2, p+1, cnt+1)) return true;
29 vis[x] = false;
30 }
31 }
32 if (d2 < n2)
33 {
34 int x = (s[p] - '0') * 10 + (s[p+1] - '0');
35 if (x >= 0 && x <= tot && !vis[x])
36 {
37 vis[x] = true; ans[cnt+1] = x;
38 if (DFS (d1, d2+1, p+2, cnt+1)) return true;
39 vis[x] = false;
40 }
41 }
42
43 return false;
44 }
45
46 int main(void) //Gym 100553J Jokewithpermutation
47 {
48 // freopen ("J.in", "r", stdin);
49 freopen ("joke.in", "r", stdin);
50 freopen ("joke.out", "w", stdout);
51
52 while (scanf ("%s", s + 1) == 1)
53 {
54 memset (vis, false, sizeof (vis));
55 len = strlen (s + 1);
56 tot = (len + 9) / 2;
57 if (tot <= 9)
58 {
59 for (int i=1; i<=len; ++i)
60 {
61 printf ("%c%c", s[i], (i==len) ? '
' : ' ');
62 }
63 }
64 else
65 {
66 n1 = 9; n2 = tot - 9;
67 DFS (0, 0, 1, 0);
68 for (int i=1; i<=tot; ++i)
69 {
70 printf ("%d%c", ans[i], (i==tot) ? '
' : ' ');
71 }
72 }
73 }
74
75 return 0;
76 }