动态规划之最长公共子序列

给定两个序列x和y，称z是x和y的公共子序列，如果z既是x的子序列，又是y的子序列；最长的公共子序列称作最长公共子序列LCS（longest common subsequence）。

解题思路

(1)LCS的最优子结构
设zk是xm和yn的一个LCS，则，如果x和y的最后一个元素相同，则z中去掉最后一个元素之后zk-1仍为xm-1和yn-1的LCS。
如果xm！=yn，若zk！=xm，则z是xm-1和y的一个LCS，若zk！=yn，则z是xm和yn-1的LCS。

(2)一个递归解
设c[i][j]为序列xi和yj的一个LCS的长度，则有：

(3)计算LCS的长度

实现代码

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int lenLCS(const char *ch1, const char *ch2, int len1, int len2, int **c)
{
    for (int i = 0; i <= len1; i++)
    {
        c[i][0] = 0;
    }
    for (int i = 0; i <= len2; i++)
    {
        c[0][i] = 0;
    }

    for (int i = 1; i <= len1; i++)
    {
        for (int j = 1; j <= len2; j++)
        {
            if (ch1[i-1] == ch2[j-1])
            {
                c[i][j] = c[i-1][j-1] + 1;
            }
            else
            {
                if (c[i-1][j] >= c[i][j-1])
                {
                    c[i][j] = c[i-1][j];
                }
                else
                {
                    c[i][j] = c[i][j-1];
                }
            }
        }
    }

//    for (int i = 0; i <= len1; i++)
//        for (int j = 0; j <= len2; j++)
//            if (j == len2) printf("%d\n", c[i][j]);
//            else printf("%d ", c[i][j]);

    return c[len1][len2];
}

void printLCS(const char *ch1, const char* ch2, int len1, int len2, int **c)
{
    int i = len1;
    int j = len2;
    while (c[i][j] > 0)
    {
        if (ch1[i-1] == ch2[j-1])
        {
            cout<<ch1[i-1];
            i--;
            j--;
        }
        else
        {
            if (c[i][j] == c[i-1][j])
            {
                i--;
            }
            else
            {
                j--;
            }
        }
    }
}

int main()
{
    char *ch1 = "ACCGGTCGAGTGCGCGGAAGCCGGCCGAA";
    char *ch2 = "GTCGTTCGGAATGCCGTTGCTCTGTAAA";
    int len1 = strlen(ch1);
    int len2 = strlen(ch2);
    int **c = new int*[len1 + 1];
    for (int i = 0; i <= len1; i++)
    {
        c[i] = new int[len2 + 1];
    }

    cout<<lenLCS(ch1, ch2, len1, len2, c)<<endl;
    printLCS(ch1, ch2, len1, len2, c);

    for (int i = 0; i <= len1; i++)
    {
        delete [] c[i];
    }
    delete [] c;
    return 0;
}

不连续情况的转移方程：

扩展到连续情况，转移方程为：

实现代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int fun(char *ch1, char *ch2, int len1, int len2, int **c)
{
    for (int i = 0; i <= len1; i++)
    {
        c[i][0] = 0;
    }
    for (int j = 0; j <= len2; j++)
    {
        c[0][j] = 0;
    }

    int maxlen = 0;
    for (int i = 1; i <= len1; i++)
    {
        for (int j = 1; j <= len2; j++)
        {
            if (ch1[i-1] == ch2[j-1])
            {
                c[i][j] = c[i-1][j-1] + 1;
            }
            else
            {
                c[i][j] = 0;
            }

            maxlen = max(maxlen, c[i][j]);
        }
    }

    return maxlen;
}

int main()
{
    char *ch1 = "acaccbabb";
    char *ch2 = "acbac";
    int len1 = strlen(ch1);
    int len2 = strlen(ch2);
    int **c = new int*[len1 + 1];
    for (int i = 0; i <= len1; i++)
    {
        c[i] = new int[len2 + 1];
    }
    int maxlen = fun(ch1, ch2, len1, len2, c);
    printf("The max length is : %d\n", maxlen);

    for (int i = 0; i <= len1; i++)
    {
        delete [] c[i];
    }
    delete [] c;
}

转自：http://blog.csdn.net/foreverling/article/details/44679139