POJ 3356. AGTC

1. 题目


Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10970 Accepted: 4217


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.


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

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 nm.

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.


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


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

Sample Input

Sample Output


2. 思路




  • 如果x[i]和y[j]相同,则 f[i][j] = f[i - 1][j - 1] ,否则将x[i]修改为y[j],在f[i – 1][j – 1]基础上增加一步操作, f[i][j] = f[i - 1][j - 1] + 1 。
  • 在x[1 : i – 1]的基础上增加一个与y[j]相同的字符, f[i][j] = f[i - 1][j] + 1 。
  • 删除x[i], f[i][j] = f[i][j - 1] + 1 。


3. 代码