PKU3356 AGTC (简单DP)

最新推荐文章于 2022-02-25 19:38:17 发布

原创最新推荐文章于 2022-02-25 19:38:17 发布 · 675 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #integer #input #output #c

本文介绍了一种通过动态规划算法来解决字符串转换问题的方法，即如何以最少的操作次数将一个字符串转换成另一个字符串。文章详细解释了算法的具体实现，并通过一个具体的示例展示了算法的应用过程。

AGTC

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4096		Accepted: 1546

Description

Let x and y be two strings over some finite alphabet A . We would like to transform x into y allowing only operations given below:

Deletion: a letter in x is missing in y at a corresponding position.
Insertion: a letter in y is missing in x at a corresponding position.
Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration
A G T A A G T * A G G C



| | |       |   |   | |



A G T * C * T G A C G C
Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C



|  |  |        |     |     |  |



A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m .

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y .

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y .

Sample Input

10 AGTCTGACGC

11 AGTAAGTAGGC

Sample Output

Source

Manila 2006

best[i][j]表示a串的前i个元素变成b串的前j个元素所要经过的最少步骤，开始时先初始化，best[i][0] = i,best[0][j] = j。其中，i和j是从0开始，但是当为1的时候才表示串的第1个元素。另外，比如best[5][0]表示有5个元素的a串变成0个元素的b串要经过的最少步骤，显然，这种时候只有把a中的元素全部删除（删5次），所以best[5][0] = 5。接着开始DP，当a[i] == b[j]，则best[i][j] = best[i - 1][j - 1];当不等时，best[i][j] = min(best[i - 1][j - 1],best[i][j - 1],best[i - 1][j]) + 1，min中三个分别表示三种操作：更改，插入，删除。最后提醒一下这题目是多case的，题目本身没说清楚。