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  • ACM 编辑距离

    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

    4

    解题思路:

    题目大意是将两个字符数组通过删除字符、插入字符、改变字符。来使这两个字符串变成相等的字符串。问我们最少需要经过多少步能使这两个字符串变成相等的字符串。

    优子结构:dp[i][j]表示从a[i]到b[j]完全匹配的最小操作数

    状态转移方程:

              1.dp[i][0]=i,dp[0][i]=i     //这是初始化步骤,这符合规律,因为这种情况下只能执行删除操作,

                     2.dp[i][j]=dp[i-1][j-1] (a[i]=b[j]) //相等无需变化,因此操作数也不增加

              3.dp[i][j]=min{dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+1} (a[i]!=b[j])  //不相等还要考虑替换,插入          操作
    程序代码:
    #include <iostream>
    using namespace std;
    char a[1005],b[1005];
    int dp[1005][1005];
    int n1,n2;
    int min(int x,int y)
    {
    	return x<y?x:y;
    }
    int maxlen(char *str1,char *str2)
    {
    	int i,j;
        //n1=strlen(str1),n2=strlen(str2);
    	int len=n1>n2?n1:n2;
        for(i=0;i<=len;i++)
            dp[i][0]=dp[0][i]=i;
        for(i=1;i<=n1;i++)
    	{
            for(j=1;j<=n2;j++)   //以下从str1[0]和str2[0]开始比较
    		{
                if(a[i]==b[j])
    				dp[i][j]=dp[i-1][j-1];
    			else
    				dp[i][j]=min( min(dp[i][j-1]+1,dp[i-1][j]+1),dp[i-1][j-1]+1);
    		}
    	}
        return dp[n1][n2];
    }
    int main()
    {
        while(~scanf("%d%s",&n1,a+1))
         {
             scanf("%d%s",&n2,b+1);
             cout<<maxlen(a,b)<<endl;
         }
        return 0;
    }
    
    
    

      

    
    
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  • 原文地址:https://www.cnblogs.com/xinxiangqing/p/4740538.html
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