[UVA] 1262 - Password
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.
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 pi appears 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.
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 5 grid 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.
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
1
K7, 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
題目描述：
一般的密碼鎖是中間一排對應道的密碼都要正確才會開啟，但是這個密碼鎖只要出現在能看到的 6x5 中的每一列中都出現密碼的對應的那個即可，也就是不用對齊。
給定兩張都具備共同密碼的圖示，可能會有很多種可能的密碼，按照字典排序後，輸出第 k-th 個
題目解法：
把兩張圖的每一列計算交集的字母，然後依序窮舉，直到個數 = K
因為 K 最多也才 7776，就可以依序窮舉，而不用使用數學計算的剪枝。
#include <stdio.h>
#include <string.h>
int board[6][26], n;
int cnt, found;
char ret[10];
void dfs(int idx) {
if(found == 1) return;
if(idx == 5) {
cnt++;
if(cnt == n) {
ret[idx] = '\0';
puts(ret);
found = 1;
}
return;
}
int i;
for(i = 0; i < 26; i++) {
if(board[idx][i] == 2) {
ret[idx] = i+'A';
dfs(idx+1);
}
}
}
int main() {
int testcase;
scanf("%d", &testcase);
while(testcase--) {
memset(board, 0, sizeof(board));
int i, j, k;
char s[6][10];
scanf("%d", &n);
for(i = 0; i < 6; i++)
scanf("%s", s[i]);
for(i = 0; i < 5; i++) {
int ss[26] = {};
for(j = 0; j < 6; j++)
ss[s[j][i]-'A'] = 1;
for(j = 0; j < 26; j++)
if(ss[j])
board[i][j]++;
}
for(i = 0; i < 6; i++)
scanf("%s", s[i]);
for(i = 0; i < 5; i++) {
int ss[26] = {};
for(j = 0; j < 6; j++)
ss[s[j][i]-'A'] = 1;
for(j = 0; j < 26; j++)
if(ss[j])
board[i][j]++;
}
cnt = 0, found = 0;
dfs(0);
if(found == 0)
puts("NO");
}
return 0;
}