UVa 1262 - Password (decode)

Link:

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3703

 

Title:

Given two alphabetic matrices of 6 rows and 5 columns, find a "password" that satisfies the following conditions: each letter in the password appears in the corresponding column of both matrices.
For example, the 2nd letter from the left must appear in the 2nd column from the left in both matrices.
Given k (1≤k≤7777), your task is to find the kth smallest lexicographical password. If not present, output NO.

 

analyze:

Classic decoding problem. First find all possible letters in each column of the answer, then decode bit by bit.
See the code for specific implementation.

 

Code:

 1 import java.io.*;
 2 import java.util.*;
 3 
 4 public class Main {
 5     static char G[][][] = new char[2][6][9];
 6     
 7     public static void main(String args[]) {
 8         Scanner cin = new Scanner(new BufferedInputStream(System.in));
 9         
10         int T = cin.nextInt();
11         while(T --> 0) {
12             intk = cin.nextInt();
 13              cin.nextLine(); // clear buffer 
14              for ( int t = 0; t < 2; t++ ) {
 15                  for ( int r = 0; r < 6; r++ ) {
 16                      G[t][r] = cin.nextLine().toCharArray();
 17                  }
 18              }
 19              
20              @SuppressWarnings("unchecked" )
 21              ArrayList<Character> a[] = new ArrayList[5]; // a [c]: All possible letters in column c of the answer 
22              for (int i = 0; i < 5; i++) a[i] = new ArrayList<Character> ();
 23              
24              boolean has[][][] = new  boolean [2][5][26]; // The first Is there a letter ch in column c of t matrices 
25              for ( int c = 0; c < 5; c++ ) {
 26                  for ( int t = 0; t < 2; t++ ) {
 27                      for ( int r = 0; r < 6; r++ ) {
 28                          int ch = G[t][r][c] - 'A' ;
 29                          has[t][c][ch] = true ;
 30                     }
31                 }
32                 for(int i = 0; i < 26; i++)
33                     if(has[0][c][i] && has[1][c][i]) a[c].add((char)(i+'A'));
34             }
35             
36             int up = 1;
37             for(int i = 0; i < 5; i++) up *= a[i].size();
38             if(k > up) {
39                 System.out.println("NO");
40                 continue;
41             }
42             
43             k--; // 解码
44             for(int i = 0; i < 5; i++) {
45                 up /= a[i].size();
46                 System.out.print(a[i].get(k/up));
47                 k %= up;
48             }
49             System.out.println();
50         }
51         cin.close();
52     }
53 }