lcs算法

1. 问题

 

 

 

2. 解析

 

 

 

3. 设计

 

 

 

 

4. 分析

 时间复杂度为O(mn)，空间复杂度为O(mn)。

5. 源码

#include <stdio.h>

#include <string.h>

#define MAXLEN 50

 

void LCSLength(char *x, char *y, int m, int n, int c[][MAXLEN], int b[][MAXLEN])

{

    int i, j;

 

    for(i = 0; i <= m; i++)

        c[i][0] = 0;

    for(j = 1; j <= n; j++)

        c[0][j] = 0;

    for(i = 1; i<= m; i++)

    {

        for(j = 1; j <= n; j++)

        {

            if(x[i-1] == y[j-1])

            {

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

                b[i][j] = 1;                 

            }                                   

            else if(c[i-1][j] >= c[i][j-1])

            {

                c[i][j] = c[i-1][j];

                b[i][j] = 3;

            }

            else

            {

                c[i][j] = c[i][j-1];

                b[i][j] = 2;

            }

        }

    }

}

 

void PrintLCS(int b[][MAXLEN], char *x, int i, int j)

{

    if(i == 0 || j == 0)

        return;

    if(b[i][j] == 1)

    {

        PrintLCS(b, x, i-1, j-1);

        printf("%c ", x[i-1]);

    }

    else if(b[i][j] == 3)

        PrintLCS(b, x, i-1, j);

    else

        PrintLCS(b, x, i, j-1);

}

 

int main()

{

    char x[MAXLEN] = {"ABCBDAB"};

    char y[MAXLEN] = {"BDCABA"};

 

    int  b[MAXLEN][MAXLEN];                       

    int  c[MAXLEN][MAXLEN];

 

    int m, n;

 

    m = strlen(x);

    n = strlen(y);

 

    LCSLength(x, y, m, n, c, b);

    PrintLCS(b, x, m, n);

 

    return 0;

}