POJ 1635 树的最小表示法

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <string>
 5 #include <algorithm>
 6 #include <vector>
 7 using namespace std;
 8 #define N 3010
 9 char s1[N] , s2[N];
10 
11 string dfs(int l , int r , char *s)
12 {
13    // cout<<l<<" "<<r<<endl;
14     vector<string> v;
15     int num = 0;
16     string ret = "";
17     for(int i=l ; i<=r ; i++){
18         if(s[i] == ‘0‘) num++;
19         else num--;
20         if(num==0){
21             string tmp = "0"+dfs(l+1 , i-1 , s)+"1";
22             v.push_back(tmp);
23             l = i+1;
24         }
25     }
26     sort(v.begin() , v.end());
27     for(int i=0 ; i<v.size() ; i++)
28         ret += v[i];
29     return ret;
30 }
31 
32 int main()
33 {
34     #ifndef ONLINE_JUDGE
35         freopen("a.in" , "r" , stdin);
36     #endif // ONLINE_JUDGE
37     int T;
38     scanf("%d" , &T);
39     while(T--){
40         scanf("%s%s" , s1 , s2);
41         string t1 = dfs(0 , strlen(s1)-1 , s1);
42         string t2 = dfs(0 , strlen(s2)-1 , s2);
43       //  cout<<t1<<" "<<t2<<endl;
44         printf("%s\n" , (t1==t2)?"same":"different");
45     }
46     return 0;
47 }