2013
11-13

# Copying DNA

Evolution is a seemingly random process which works in a way which resembles certain approaches we use to get approximate solutions to hard combinatorial problems. You are now to do something completely different.

Given a DNA string S from the alphabet {A,C,G,T}, find the minimal number of copy operations needed to create another string T. You may reverse the strings you copy, and copy both from S and the pieces of your partial T. You may put these pieces together at any time. You may only copy contiguous parts of your partial T, and all copied strings must be used in your final T. Example: From S = “ACTG” create T = “GTACTATTATA”

1. Get GT……… by copying and reversing "TG" from S.
2. Get GTAC……. by copying "AC" from S.
3. Get GTAC…TA.. by copying "TA" from the partial T.
4. Get GTAC…TAAT by copying and reversing "TA" from the partial T.
5. Get GTACAATTAAT by copying "AAT" from the partial T.

The first line of input gives a single integer, 1 ≤ t ≤ 100, the number of test cases. Then follow, for each test case, a line with the string S of length 1 ≤ m ≤ 18, and a line with the string T of length 1 ≤ n ≤ 18.

Output for each test case the number of copy operations needed to create T from S, or "impossible" if it cannot be done.

5
ACGT
GTAC
A
C
ACGT
TGCA
ACGT
TCGATCGA
A
AAAAAAAAAAAAAAAAAA

2
impossible
1
4
6

//* @author
/**
* @version 2009/08/26
* @author sbzlyessit
*/

/*搜索题. 状态不多, 可用记忆化搜索, 但是转移耗费巨大, 所以仍需剪枝.
*最主要的一项剪枝就是每一位匹配时都要取最长的长度进行下一步搜索.
*/

import java.io.*;
import java.util.*;

public class Main {

private static final int        MAX_N = 18;

private static char[]           s, t, sr;
private static int[]            matchs, matchsr;
private static int[]            hash = new int[1 <<  MAX_N];

public static void main(String[] argv) throws Exception {
for(int caseSize = Integer.parseInt(in.readLine()); caseSize > 0; caseSize--){
System.out.println("impossible");
} else {
System.out.println(solve(0));
}
}
}

private static boolean readIn() throws Exception {
int     i;
if ((!S.contains("A") && T.contains("A")) || (!S.contains("C") && T.contains("C")) ||
(!S.contains("G") && T.contains("G")) || (!S.contains("T") && T.contains("T"))) {
return false;
}
s = S.toCharArray();
t = T.toCharArray();
sr = new char[s.length];
matchs = new int[t.length];
matchsr = new int[t.length];
Arrays.fill(hash, Integer.MAX_VALUE);
hash[(1 <<  t.length) - 1] = 0;
reverse(s, sr);
for (i = 0; i <  t.length; i++) {
matchs[i] = longestMatch(i, t.length, s);
matchsr[i] = longestMatch(i, t.length, sr);
}
return true;
}

private static void reverse(char[] a, char[] ar) {
for (int i = 0; i < a.length; i++) {
ar[i] = a[a.length - i - 1];
}
}

private static int longestMatch(int from, int to, char[] a) {
int i, len, max = 0;
for (i = 0, len = 0; i < a.length; i++, len = 0) {
while (len < to - from && i + len < a.length && a[i + len] == t[from + len]) {
len++;
}
max = Math.max(max, len);
}
return max;
}

private static int solve(int state) {
if (hash[state] != Integer.MAX_VALUE) {
return hash[state];
}
char[] done = new char[t.length];
char[] doner = new char[t.length];
int i, m = 0, len, g;
for (i = 0; i < done.length; i++) {
done[i] = (state & (1 << i)) != 0 ? t[i] : '0';
}
reverse(done, doner);
for(i = 0; i < t.length; i++) {
if ((state & (1 << i)) != 0) {
m = 0;
continue;
}
len = 1;
while (i + len < t.length && (state & (1 << (i + len))) == 0) {
len++;
}
g = Math.min(len, Math.max(matchs[i], matchsr[i]));
g = Math.max(g, longestMatch(i, i + len, done));
g = Math.max(g, longestMatch(i, i + len, doner));
if(g > m - 1) {
hash[state] = Math.min(hash[state], 1 + solve(state | ((1 << (i + g)) - (1 << i))));
}
m = g;
}
return hash[state];
}

}

1. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮