UVA 1262 Password 暴力枚举

Password

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA.
Original ID: 1262

64-bit integer IO format: %lld     
Java class name: Main

pid=18400" class="btn" style="">Prev

Submit 

showpid=18401" class="btn" style="">Status 

pid=18401" class="btn" style="">Statistics Discuss

Next

[PDF Link]

Shoulder-surfing is the behavior of intentionally and stealthily watching the screen of another person's electronic device, such as laptop computer or mobile phone. Since mobile devices prevail, it is getting serious to steal personal information by shoulder-surfing.

Suppose that we have a smart phone. If we touch the screen keyboard directly to enter the password, this is very vulnerable since a shoulder-surfer easily knows what we have typed. So it is desirable to conceal
the input information to discourage shoulder-surfers around us. Let me explain one way to do this.

You are given a 6 x 5 grid. Each column can be considered the visible part of a wheel. So you can easily rotate
each column wheel independently to make password characters visible. In this problem, we assume that each wheel contains the 26 upper letters of English alphabet. See the following Figure 1.

Figure 1. 6 x 5 window clips a valid grid representation for a password.

Assume that we have a length-5 password such as p1 p2 p3 p4 p5.
In order to pass the authentication procedure, we should construct a configuration of grid space where each piappears
in the i-th column of the grid. In that situation we say that the user password is accepted.

Let me start with one example. Suppose that our password was set `COMPU'. If we construct the grid as shown in Figure 2 on next page, then the authentication is successfully processed.

Figure 2. A valid grid representation for password `COMPU'.

In this password system, the position of each password character in each column is meaningless. If each of the 5 characters in p1 p2 p3 p4 p5 appears
in the corresponding column, that can be considered the correct password. So there are many grid configurations allowing one password. Note that the sequence of letters on each wheel is randomly determined for each trial and for each column. In practice, the
user is able to rotate each column and press ``Enter" key, so a should-surfer cannot perceive the password by observing the 6 x 5grid since there are too
many password candidates. In this 6 x 5 grid space, maximally 65 =
7, 776 cases are possible. This is the basic idea of the proposed password system against shoulder-surfers.

Unfortunately there is a problem. If a shoulder-surfer can observe more than two grid plate configurations for a person, then the shoulder-surfer can reduce the searching space and guess the correct password. Even
though it is not easy to stealthily observe other's more than once, this is one weakness of implicit grid passwords.

Let me show one example with two observed configurations for a grid password. The user password is `COMPU', but `DPMAG' is also one candidate password derived from the following configuration.

Figure 3. Both of `COMPU' and `DPMAG' are feasible password .

You are given two configurations of grid password from a shoulder-surfer. Suppose that you have succeeded to stealthily record snapshots of the target person's device (e.g. smart phone). Then your next task is to
reconstruct all possible passwords from these two snapshots. Since there are lots of password candidates, you are asked for the k-th password among
all candidates in lexicographical order. In Figure 3, let us show the first 5 valid password. The first 5 valid passwords are `ABGAG' , `ABGAS', `ABGAU', `ABGPG' and `ABGPS'.

The number k is given in each test case differently. If there does not exist a k-th
password since k is larger than the number of all possible passwords, then you should print `NO' in the output.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is
given in the first line of the input. The first line of each test case contains one integer, K, the order of the password you should find. Note that 1K7,
777. Next the following 6 lines show the 6 rows of the first grid and another 6 lines represent the 6 rows of the second grid.

Output

Your program is to write to standard output. Print exactly the k-th password (including `NO') in one line for each test case.

The following shows sample input and output for three test cases.

Sample Input

3
1
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
5
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
64
FGHIJ
EFGHI
DEFGH
CDEFG
BCDEF
ABCDE
WBXDY
UWYXZ
XXZFG
YYFYH
EZWZI
ZGHIJ

Sample Output

ABGAG
ABGPS
NO

pid=18400" class="btn" style="">Prev

Submit Status Statistics 

pid=18401" class="btn" style="">Discuss

Next

---

Distributed under GPLv3. | Project Homepage | Developer: 51isoft |
Current Style: Cerulean.

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月15日 星期一 10时08分57秒
File Name     :UVA1262.cpp
 ************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>

using namespace std;

int k;
char m1[10][10],m2[10][10];
vector<char> vc[10];
vector<string> vs;
char str[10];

void dfs(int deep)
{
	if(deep==5)
	{
		string ans;
		for(int i=0;i<5;i++)
			ans+=str[i];
		vs.push_back(ans);
		return ;
	}
	for(int i=0,sz=vc[deep].size();i<sz;i++)
	{
		str[deep]=vc[deep][i];
		dfs(deep+1);
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&k);
		for(int i=0;i<6;i++)
			scanf("%s",m1[i]);
		for(int i=0;i<6;i++)
			scanf("%s",m2[i]);

		for(int j=0;j<5;j++)
		{
			vc[j].clear();
			for(int i=0;i<6;i++)
			{
				char c1=m1[i][j];
				bool flag=false;
				for(int i2=0;i2<6&&flag==false;i2++)
				{
					if(m2[i2][j]==c1) flag=true;
				}
				if(flag==true)
				{
					vc[j].push_back(c1);
				}
			}
		}

		vs.clear();
		dfs(0);
		sort(vs.begin(),vs.end());
		int tt=unique(vs.begin(),vs.end())-vs.begin();
		if(k>tt) puts("NO");
		else
		{
			cout<<vs[k-1]<<endl;
		}
	}

	return 0;
}