Unshuffle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 148    Accepted Submission(s): 43
Special Judge

Problem Description A shuffle of two strings is formed by interspersing the characters into a new string, keeping the characters of each string in order. For example, MISSISSIPPI is a shuffle of MISIPP and SSISI. Let me call a string square if it is a shuffle of two identical strings. For example, ABCABDCD is square, because it is a shuffle of ABCD and ABCD, but the string ABCDDCBA is not square.

Given a square string, in which each character occurs no more than four times, unshuffle it into two identical strings.  

Input First line, number of test cases, T.

Following are 2*T lines. For every two lines, the first line is n, length of the square string; the second line is the string. Each character is a positive integer no larger than n.

T<=10, n<=2000.  

Output T lines. Each line is a string of length n of the corresponding test case. ‘0’ means this character belongs to the first string, while ‘1’ means this character belongs to the second string. If there are multiple answers, output any one of them.  

Sample Input 1 8 1 2 3 1 2 4 3 4  

Sample Output 00011011  

Source
2013 Multi-University Training Contest 6  

Recommend zhuyuanchen520           题解说的2-SAT，虽然比赛一直往这方面想，但是构图一直没有想出来。     只会暴力搜索过去。   T_T   dfs标记下就行了  

 1 /*
 2  * Author:  kuangbin
 3  * File Name: 1011.cpp
 4  */
 5 #include <iostream>
 6 #include <stdio.h>
 7 #include <string.h>
 8 #include <algorithm>
 9 using namespace std;
10 
11 const int MAXN = 2020;
12 int a[MAXN];
13 int s[MAXN];
14 char ans[MAXN];
15 
16 bool flag ;
17 int n;
18 void dfs(int cur,int t1,int t2)
19 {
20     if(flag)return;
21     if(t1 > n/2 || t2 > n/2)return;
22     if(cur == n)
23     {
24         flag = true;
25         return;
26     }
27     s[t1] = a[cur];
28     ans[cur] = '0';
29     dfs(cur+1,t1+1,t2);
30     if(flag)return;
31     if(s[t2] == a[cur])
32     {
33         ans[cur] = '1';
34         dfs(cur+1,t1,t2+1);
35     }
36 }
37 
38 int main()
39 {
40     int T;
41     scanf("%d",&T);
42     while(T--)
43     {
44         scanf("%d",&n);
45         for(int i = 0;i < n;i++)
46             scanf("%d",&a[i]);
47         flag = false;
48         dfs(0,0,0);
49         for(int i = 0;i < n;i++)printf("%c",ans[i]);
50         printf("\n");
51     }
52     return 0;
53 }