HDU 1158 Common Subsequence-优快云博客

本文介绍了一个经典的动态规划问题——寻找两个字符串的最长公共子序列。通过输入两个字符串，程序能够找出它们之间的最长公共子序列长度。文章提供了一段C语言实现的代码示例，该算法使用二维矩阵存储中间结果，避免了重复计算。

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Problem Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

这是一个动态规划入门题
if(i==0||j==0)matrix[i][j]=0;
if(s1[i-1]==s2[j-1])matrix[i][j]=matrix[i-1][j-1]+1;
if(s1[i-1]!=s2[j-1])matrix[i][j]=max(matrix[i-1][j],matrix[i][j-1]);

#include<stdio.h>
#include<string.h>
#define max 505
int main()
{
    char s1[max],s2[max];
    int matrix[max][max],len1,len2;
    int i,j;
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        len1=strlen(s1);
        len2=strlen(s2);
        for(i=0; i<=len1; i++)matrix[i][0]=0;
        for(j=0; j<=len2; j++)matrix[0][j]=0;
        for(i=1; i<=len1; i++)
        {
            for(j=1; j<=len2; j++)
            {
                if(s1[i-1]==s2[j-1])
                {
                    matrix[i][j]=matrix[i-1][j-1]+1;
                }
                else
                {
                    if(matrix[i-1][j]>=matrix[i][j-1])
                    {
                        matrix[i][j]=matrix[i-1][j];
                    }
                    else
                    {
                        matrix[i][j]=matrix[i][j-1];
                    }
                }
            }
        }
        printf("%d\n",matrix[len1][len2]);
    }
    return 0;
}