POJ 3356 AGTC（dp之公共最长子序列）

偏执的太偏执、 2022-06-10 23:08 174阅读 0赞

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

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

题解：

很简单的一题，就是找公共最长子序列，然后用最长的串减去公共子序列长度就是答案。。练习一下公共最长子序列

代码：

#include<iostream>
    #include<stdio.h>
    #include<map>
    #include<algorithm>
    #include<math.h>
    #include<queue>
    #include<stack>
    #include<string>
    #include<cstring>
    using namespace std;
    char s1[1005],s2[1005];
    int dp[1005][1005];
    int main()
    {
        int i,j,k,n,m;
        while(scanf("%d%s%d%s",&n,s1,&m,s2)!=EOF)
        {
            dp[0][0]=0;
            dp[0][1]=0;
            dp[1][0]=0;
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    if(s1[i-1]==s2[j-1])
                    {
                        dp[i][j]=dp[i-1][j-1]+1;
                    }
                    else
                    {
                        dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                    }
                }
            }
            printf("%d\n",max(n,m)-dp[n][m]);
        }
        return 0;
    }