POJ 3356 AGTC 最短编辑距离

AGTC

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12261 Accepted: 4602

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 AGTCTGACGC11 AGTAAGTAGGC

Sample Output

4

Source

Manila 2006

点击打开题目链接

由字符串 A 通过下列三种操作

　1、插入一个字符；

   2、删除一个字符；

　3、改变一个字符

变换到字符串 B 所需要的最少操作次数（感觉和最长公共子序列有点相似）

状态转移方程：

dp[i][j] 表示字符串A判断到位置 i ，B判断到位置j需要的最少次数。

一：A[i] == B[j] 此时 dp[i][j] = dp[i - 1][j - 1] 

二：A[i] ！= B[j] 此时有三种情况可以导致我们上面设计的状态会发生转移。
①、我们可以在 B[j]后面插入一个核苷酸（即一个字符）ch，ch==A[i]，这样做的话，
至少需要 dp[i - 1][j] + 1步操作，即 dp[i][j] = dp[i - 1][j] + 1。
②、我们可以删除 B[j]，这样的话，B[1...j] 变为A[1...i] 需要 dp[i][j - 1]步，
即 dp[i][j] = dp[i][j - 1] + 1。
③、我们也可以考虑修改 B[j]，使它变为A[i]，那修改相当于用了 1 步。
所以 dp[i][j] = dp[i - 1][j - 1] +  1。

从上面三种状态转移中选择一个最小值就可以了。

处理边界：
处理好边界非常重要，这里需要注意的是对dp[0][0....m],dp[0.....n][0]的初始化，
可以这样看，dp[0][i],就是说A[1...n]是一个空串，而B[1...m]是个长度为i的串，
很显然B串变为A串就是删除i个核苷酸。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int maxn = 1000 + 5;char A[maxn], B[maxn];int dp[maxn][maxn];int len1, len2;void init(){memset(dp, 0, sizeof(dp));int Len = max(len1, len2);for (int i = 0; i <= Len; i++)//处理边界 dp[i][0] = dp[0][i] = i;}int solve(){for (int i = 0; i < len1; i++){for (int j = 0; j < len2; j++)//状态转移，选三种状态中最小的 {if (A[i] == B[j])dp[i + 1][j + 1] = dp[i][j];else dp[i + 1][j + 1] = min(dp[i][j] + 1, min(dp[i + 1][j] + 1, dp[i][j + 1] + 1));}}return dp[len1][len2];}int main(){while (~scanf("%d%s", &len1, A)){scanf("%d%s", &len2, B);init();printf("%d\n", solve());}    return 0;}