就是一个文件的很多副本碎成两半， 让你根据所有碎片求出原来的文件
排序枚举一下就ok了
题目：
Question 2: File Fragmentation
The Problem
Your friend, a biochemistry major, tripped while carrying a tray of computer files through the lab. All of the files fell to the ground and broke. Your friend picked up all the file fragments and called you to ask for help putting them back together again.
Fortunately, all of the files on the tray were identical, all of them broke into exactly two fragments, and all of the file fragments were found. Unfortunately, the files didn't all break in the same place, and the fragments were completely mixed up by their fall to the floor.
You've translated the original binary fragments into strings of ASCII 1's and 0's, and you're planning to write a program to determine the bit pattern the files contained.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
Input will consist of a sequence of ``file fragments'', one per line, terminated by the end-of-file marker. Each fragment consists of a string of ASCII 1's and 0's.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
Output is a single line of ASCII 1's and 0's giving the bit pattern of the original files. If there are 2N fragments in the input, it should be possible to concatenate these fragments together in pairs to make N copies of the output string. If there is no unique solution, any of the possible solutions may be output.
Your friend is certain that there were no more than 144 files on the tray, and that the files were all less than 256 bytes in size.
Sample Input
1 011 0111 01110 111 0111 10111
Sample Output
01110111
代码：
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <algorithm> 5 #include <memory.h> 6 #include <set> 7 using namespace std; 8 9 10 string frag[200]; 11 int p; 12 int vis[200]; 13 int len; 14 set<string> v; 15 int judge(string str) 16 { 17 // cout<<"str : "<<str<<endl; 18 for(int i=0;i<p;i++) 19 { 20 for(int j=p-1;j>i;j--) 21 { 22 if(!vis[i] && !vis[j] && frag[i]+frag[j]==str || frag[j]+frag[i]==str) 23 { 24 vis[i]=vis[j]=1; 25 } 26 } 27 if(!vis[i]) 28 { 29 // cout<<" I "<<i << " frag :" <<frag[i]<<endl; 30 return 0; 31 } 32 } 33 if(!v.count(str)) 34 { 35 v.insert(str); 36 37 cout<<str<<endl; 38 } 39 return 1; 40 } 41 42 int solve() 43 { 44 v.clear(); 45 memset(vis,0,sizeof(vis)); 46 string tmp ; 47 48 len = frag[0].length()+ frag[p-1].length(); 49 50 for(int i=0;i<p;i++) 51 { 52 53 for(int j=p-1;j>i;j--) 54 { 55 int tlen = frag[i].length()+frag[j].length(); 56 if(tlen<len) 57 { 58 break; 59 } 60 61 vis[i]=1;vis[j]=1; 62 63 judge (frag[i]+frag[j] ); 64 memset(vis,0,sizeof(vis)); 65 judge(frag[j]+frag[i]); 66 memset(vis,0,sizeof(vis)); 67 } 68 69 } 70 71 } 72 73 bool cmp(string a, string b) 74 { 75 return a.length()<b.length(); 76 } 77 78 int main() 79 { 80 freopen("in","r",stdin); 81 // freopen("out","w",stdout); 82 83 int tst; 84 cin>>tst; 85 cin.ignore(); 86 string tmp; 87 88 89 getline(cin,tmp); 90 while(tst--) 91 { 92 p=0; 93 while(getline(cin,tmp)) 94 { 95 if(!tmp.empty()) 96 { 97 frag [p++] = tmp; 98 } 99 else 100 break; 101 } 102 103 sort(frag,frag+p,cmp); 104 solve(); 105 if(tst)cout<<endl; 106 107 } 108 109 110 return 0; 111 }