这道题作为300分的水题，直接暴力就可以过了，题意是给你一个01串s，可以通过t[i] =s[i-1] + s[i]+ s[i+1]，得到一个新串t，
其中t[0] = s[0] + s[1],t[n-1] = t[n-1]+t[n-2]。我们现在输入的是串t，要求出串s。
下面贴代码:
#include <iostream>
#include <vector>
#include <string>
#include <cstdlib>
using namespace std;
class BinaryCode
{
private:
vector<string> data[50];
public:
vector<string> decode(string val)
{
memset(data,0,sizeof(data));
inits(val);
string res;
vector<string> buf;
int k;
for(int i=0;i<data[0].size();i++)
{
res = data[0][i];
for(int j=1;j<val.length();j++)
{
for(k=0;k<data[j].size();k++)
if(cmp(res,data[j][k]))
{
res.insert(res.length(),1,data[j][k][2]);
break;
}
if(k == data[j].size())
break;
}
// cout << res << endl;
if(res.length() == val.length()+2)
buf.push_back(res);
}
if(buf.size() == 0)
buf.push_back("NONE");
if(buf.size() == 1)
buf.push_back("NONE");
if(buf[0][1] != '1' && buf[1][1] != '0')
{
if(buf[0][0] != 'N')
buf[0] = buf[0].substr(1,val.length());
if(buf[1][0] != 'N')
buf[1] = buf[1].substr(1,val.length());
return buf;
}
else
{
string temp = buf[0];
buf[0] = buf[1];
buf[1] = temp;
}
if(buf[0][0] != 'N')
buf[0] = buf[0].substr(1,val.length());
if(buf[1][0] != 'N')
buf[1] = buf[1].substr(1,val.length());
return buf;
}
void inits(string val)
{
for(int i=0;i<val.length();i++)
if(val[i] == '0')
this->data[i].push_back("000");
else if(val[i] == '1')
{
if(i!=val.length()-1)
data[i].push_back("001");
data[i].push_back("010");
if(i!=0)
data[i].push_back("100");
}
else if(val[i] == '2')
{
if(i!=0)
data[i].push_back("110");
if(i!=val.length()-1 && i != 0)
data[i].push_back("101");
if(i!=val.length()-1)
data[i].push_back("011");
}
else if(i!=val.length()-1 && i!=0)
data[i].push_back("111");
}
bool cmp(string s1,string s2)
{
if(s1[s1.length()-2] == s2[0] && s1[s1.length()-1] == s2[1])
return true;
return false;
}
};
/*
int main()
{
BinaryCode a;
string s;
cin >> s;
vector<string> t = a.decode(s);
cout << t[0] << " " << t[1] << endl;
return 0;
}
*/
