10405 - Longest Common Subsequence

Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:

abcdgh

aedfhr

is adh of length 3.

Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4e

zz1yy2xx3ww4vv

abcdgh

aedfhr

abcdefghijklmnopqrstuvwxyz

a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0

abcdefghijklmnzyxwvutsrqpo

opqrstuvwxyzabcdefghijklmn

Output for the sample input

4

3

26

14

其实思路还是很简单的，就是边界的问题有点糊涂了，看看别人的文章突然醒悟过来，囧完了=。=

还有那个超时问题，以前也写个while (1)再跳出来的题，呜呜，被笑话了，最后老老实实改了，这题写得我心酸酸的。

LCS(s1, s2) =

 { max( LCS(sub1, s2), LCS(s1, sub2) ) , when e1 != e2

 { LCS(sub1, sub2) + e1                , when e1 == e2

这个挺重要的，我好像还没有学会推出这个式子。。明天再战！

#include<iostream>

#include<cstring>

#include<cstdio>

#define max(x,y) x>y?x:y;

using namespace std;

char str1[1050];

char str2[1050];

int  a[1050][1050];//保存数

int main()

{

   int i,j;

   int len1,len2;

   while (gets(str1))

   {

       gets(str2);

       len1 = strlen(str1);

       len2 = strlen(str2);

       for (i = 1; i <= len1; i++)

       {

        for (j = 1; j <= len2; j++)

        {

            if (str1[i-1] == str2[j-1])

            {

                a[i][j] = a[i-1][j-1]+1;

            }

            else

            {

                a[i][j] = max(a[i-1][j],a[i][j-1]);

            }

        }

      }

        cout << a[len1][len2] << endl;

   }

    return 0;

}

 

当我们只要求出lcs的长度而不需要打印出lcs时，我们还可以更加节省记录的空间。事实上，我们需要知道当前位置及其左上，左方，上方的值即可，也就是我们只要2行的二维数组即可（此时的计算顺序是每行从左到右，一行一行一次算），再交替使用这个二维数组即可。

代码如下：

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

char str1[1050];

char str2[1050];

int  a[1050][1050];

int  prev[2][1050];

int main()

{

    int len1,len2,i,j;

    while (gets(str1))

    {

        gets(str2);

        len1 = strlen(str1);

        len2 = strlen(str2);

        for (i = 0; i < 2; i++)

        {

            for (j = 0; j < len2; j++)

            {

                a[i][j] =  0;

            }

        }

        for (i = 1; i <= len1; i++)//行

        {

            for (j = 1; j <= len2; j++)//列

            {

                if (str1[i-1] == str2[j-1])

                {

                    a[2][j] = a[2][j-1]+1;

                    a[1][j] = a[2][j];

                }

                else

                {

                   a[2][j] = max(a[2][j-1],a[1][j]);

                   a[1][j] = a[2][j];

                }

            }

        }

        cout << a[2][len2] << endl;

    }

    return 0;

}

如果要把lcs打印出来，我们就用一个二维数组回溯记录是从什么方向来的。

完整代码如下：

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

char str1[1050];

char str2[1050];

int  a[1050][1050];

int  prev[1050][1050];

void print_lcs(int i,int j)

{

    if (!i || !j)

    {

        return;

    }

    if (prev[i][j] == 1)

    {

        print_lcs(i-1,j-1);

        cout << str1[i-1];

    }

    if (prev[i][j] == 2)

    {

        print_lcs(i-1,j);

    }

    if (prev[i][j] == 3)

    {

        print_lcs(i,j-1);

    }

}

int main()

{

    int len1,len2,i,j;

    while (gets(str1))

    {

        gets(str2);

        len1 = strlen(str1);

        len2 = strlen(str2);

        for (i = 1; i <= len1; i++)

        {

            for (j = 1; j <= len2; j++)

            {

                if (str1[i-1] == str2[j-1])

                {

                    a[i][j] = a[i-1][j-1]+1;

                    prev[i][j] = 1;

                }

                else

                {

                    if (a[i-1][j] >= a[i][j-1])

                    {

                        a[i][j] = a[i-1][j];

                        prev[i][j] = 2;

                    }

                    else

                    {

                        a[i][j] = a[i][j-1];

                        prev[i][j] = 3;

                    }

                }

            }

        }

        cout << a[len1][len2] << endl;

        cout << "lcs为"<< endl;

        print_lcs(len1,len2);

        cout << endl;

    }

    return 0;

}