Algorithm

USACO Section6.3 Cryptcowgraphy (cryptcow)

Problem Summary

Given a string T and an encrypted string S, determine whether S can be decoded as T.

The encryption of messege S in this problem goes like:
First, insert three letters C, O, and W, one after another and in random location in S.
Second, swap the part of S between C and O, and the part between O and W.

Note that this kind of encryption can be applied several times, and T does not contain any C, O, or W.

Solution

This problem can be solved with DFS, but it certainly needs some optimization.

First, we can examine the input S and count the numbers of all characters. If the numbers of C, O, and W are not the same, or the number of any other character in S is different from that in T, then S can not be decoded as T.

Note that, the string is cut into several sequences by C, O, and W. These sequences must be in T. If not, there is no need to continue the search.

Additionally, a tip about coding.
The C++ function ”memcpy“ requires the destination and source arrays to be seperate,i.e. not overlapping. For overlapping memory blocks, “memmove” is a safer approach.

Code

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#include <iostream>
#include <fstream>
#include <cstring>
using namespace std;
const int maxn = 100;
const char t[] = "Begin the Escape execution at the Break of Dawn";
char s[maxn],backup[maxn],tmp1[maxn],tmp2[maxn];
int n,s_count[300],t_count[300];
bool match;
void swap_s(int a,int b,int c,int len)
{
	memmove(tmp1,s+a+1,b-1-a);
	memmove(tmp2,s+b+1,c-1-b);
	memmove(s+a,tmp2,c-1-b);
	memmove(s+a+c-1-b,tmp1,b-1-a);
	memmove(s+c-2,s+c+1,len-c-1);
	s[len-3]=0;
}
bool my_strstr(char *p,int l)
{
	bool flag;
	for(int i=0;i<47;i++)
		if(t[i]==p[0])
		{
			flag=true;
			for(int j=0;j<l;j++)
				if(t[i+j]!=p[j])
				{
					flag=false;
					break;
				}
			if(flag) return true;
		}
	return false;
}
void dfs(int len,int num)
{
	if(match || len<47) return;
	if(num==0)
	{
		if(strcmp(s,t)==0)
			match=true;
		
		return;
	}
	int pos[3][10],idx[30];
	pos[0][0]=pos[1][0]=pos[2][0]=idx[0]=0;
	for(int i=0;i<len;i++)
		if(s[i]=='O')
		{
			pos[0][0]++;
			pos[0][pos[0][0]]=i;
			idx[0]++;
			idx[idx[0]]=i;
		}
		else
		if(s[i]=='C')
		{
			pos[1][0]++;
			pos[1][pos[1][0]]=i;
			idx[0]++;
			idx[idx[0]]=i;
		}
		else
		if(s[i]=='W')
		{
			pos[2][0]++;
			pos[2][pos[2][0]]=i;
			idx[0]++;
			idx[idx[0]]=i;
		}
	if(pos[1][1]>pos[0][1] || pos[1][1]>pos[2][1] || pos[2][num]<pos[0][1])
		return;
	if(pos[1][1]>0 && strncmp(s,t,pos[1][1])!=0) /* first segment */
		return;
	if(pos[2][1]<len && strcmp(s+pos[2][num]+1,t+48-len+pos[2][num])!=0) /* last segment */
		return;
	for(int i=0;i<idx[0];i++)
		if(idx[i]+1<idx[i+1])
		{
			if(!my_strstr(s+idx[i]+1,idx[i+1]-idx[i]-1))
				return;
		}
	char backup[maxn];
	memcpy(backup,s,n+1);
	
	for(int i=1;i<=num;i++)
		for(int j=1;j<=num;j++)
			if(pos[0][i]>pos[1][j])
				for(int k=num;k>=1;k--)
				{
					if(pos[2][k]<=pos[0][i]) break;
					swap_s(pos[1][j],pos[0][i],pos[2][k],len);
					dfs(len-3,num-1);
					memcpy(s,backup,n+1);
				}
}
int main()
{
	fstream fin("cryptcow.in",ios::in);
	fstream fout("cryptcow.out",ios::out);
	string s1;
	getline(fin,s1);
	fin.close();
	n=s1.length();
	for(int i=0;i<n;i++)
		s[i]=s1[i];
	s[n]=0;
	for(int i=0;i<n;i++)
		s_count[s[i]]++;
	if(s_count['C']!=s_count['O'] || s_count['O']!=s_count['W'] || s_count['C']!=s_count['W'] || n-3*s_count['C']!=47)
	{
		fout<<"0 0"<<endl;
		fout.close();
		return 0;
	}
	if(s_count['C']==0)
	{
		if(strcmp(s,t)==0)
			fout<<"1 0"<<endl;
		else
			fout<<"0 0"<<endl;
		fout.close();
		return 0;		
	}
	for(int i=0;i<47;i++) t_count[t[i]]++;
	bool flag=true;
	for(int i=0;i<255;i++)
	{
		if(i=='C' || i=='O' || i=='W')
			continue;
		if(s_count[i]!=t_count[i])
		{
			flag=false;
			break;
		}
	}
	if(flag)
		dfs(n,s_count['C']);
	if(match)
		fout<<"1 "<<s_count['C']<<endl;
	else
		fout<<"0 0"<<endl;
	fout.close();
	return 0;
}