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1 /*
2 stack 容器的应用: 要求字典序升序输出,所以先搜索入栈的
3 然后逐个判断是否满足答案,若不满足,回溯继续搜索,输出所有符合的结果
4 */
5 #include <cstdio>
6 #include <iostream>
7 #include <algorithm>
8 #include <stack>
9 #include <cmath>
10 #include <cstring>
11 #include <vector>
12 using namespace std;
13
14 const int MAXN = 1e4 + 10;
15 const int INF = 0x3f3f3f3f;
16
17 string s1, s2;
18 int len;
19 stack<char> S;
20 vector<char> V;
21
22 void DFS(int push, int pop)
23 {
24 if (push == len && pop == len)
25 {
26 for (int i=0; i<V.size (); ++i)
27 cout << V[i] << " ";
28 cout << endl;
29 }
30
31 if (push + 1 <= len) //入栈
32 {
33 S.push (s1[push]);
34 V.push_back (‘i‘);
35 DFS (push+1, pop);
36 S.pop ();
37 V.pop_back ();
38 }
39
40 if (pop + 1 <= push && pop + 1 <= len && S.top () == s2[pop]) //出栈
41 {
42 char ch = S.top ();
43 S.pop ();
44 V.push_back (‘o‘);
45 DFS (push, pop+1);
46 S.push (ch);
47 V.pop_back ();
48 }
49 }
50
51 int main(void) //ZOJ 1004 Anagrams by Stack
52 {
53 //freopen ("ZOJ_1004.in", "r", stdin);
54
55 while (cin >> s1 >> s2)
56 {
57 V.clear ();
58 while (!S.empty ()) S.pop ();
59
60 len = s1.length ();
61
62 cout << "[" << endl;
63 DFS (0, 0);
64 cout << "]" << endl;
65 }
66
67 return 0;
68 }
69
70 /*
71 [
72 i i i i o o o i o o
73 i i i i o o o o i o
74 i i o i o i o i o o
75 i i o i o i o o i o
76 ]
77 [
78 i o i i i o o i i o o o
79 i o i i i o o o i o i o
80 i o i o i o i i i o o o
81 i o i o i o i o i o i o
82 ]
83 [
84 ]
85 [
86 i i o i o i o o
87 ]
88 */
stack+DFS ZOJ 1004 Anagrams by Stack
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原文地址:http://www.cnblogs.com/Running-Time/p/4442585.html