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  • poj 3356

    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 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.

    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
    题意:
    求由字符串s1,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
    变换的字符s2所须要 的最小操作次数。
    思路:这是一个求编辑最短距离问题。利用动态规划,列出状态方程,设dp[i][j]表示字符串x[1...i]和字符串y[1...j]的最短编辑距离当x[i] == y[j]时,i和j不须要编辑,要么删除,要么插入。要么替换dp[i][j] = min(dp[i-1][j-1], dp[i-1][j] + 1, dp[i][j - 1] + 1)当x[i] != y[i]时, i和j不须要编辑dp[i][j] = min(dp[i-1][j-1] + 1, dp[i-1][j] + 1, dp[i][j-1] + 1);注意初始化dp[i][0] = dp[0][i] = i;
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int maxn = 1010;
    char strx[maxn], stry[maxn];
    int lenx, leny, dp[maxn][maxn];
    int main()
    {
    
        while( scanf("%d %s", &lenx, strx  + 1) != EOF)
        {
            scanf("%d %s", &leny, stry + 1);
            int maxv = max(lenx, leny);
            dp[0][0] = 0;
            for(int i = 1; i <= maxv; i++)
                dp[0][i] = dp[i][0] = i;
            for(int i = 1; i <= lenx; i++)
            {
                for(int j = 1; j <= leny; j++)
                {
                    dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1);
                    if(strx[i] == stry[j])
                        dp[i][j] = min(dp[i][j], dp[i-1][j-1]);
                    else
                        dp[i][j] = min(dp[i][j], dp[i-1][j-1] + 1);
                }
            }
            printf("%d
    ", dp[lenx][leny]);
        }
    
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/cxchanpin/p/7079999.html
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