HDU 4665 Unshuffle(2013多校6 1011)

Unshuffle
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 148    Accepted Submission(s): 43Special 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
 
 
 
 
 
問題解説の2-SAT、試合はずっとこの方面に考えていますが、構図はずっと考えていません.
 
 
暴力的に過去を捜索するだけだ.
 
T_T
 
dfsタグでいい
 

 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("
");
51     }
52     return 0;
53 }