九度OJ 1042 Coincidence -- 动态规划（最长公共子序列）

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于 2014-02-09 15:45:41 发布

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分类专栏：九度OJ 考研机试动态规划文章标签：九度OJ 考研机试动态规划最长递增子序列

本文链接：https://blog.youkuaiyun.com/JDPlus/article/details/19010109

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本文介绍了一个解决最长公共子序列问题的算法实现。通过动态规划的方法，该程序能够找出两个字符串之间的最长公共子序列，并输出其长度。文章提供了一个完整的C语言程序示例，用于处理长度不超过100的字符串。

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题目地址：http://ac.jobdu.com/problem.php?pid=1042

题目描述：

Find a longest common subsequence of two strings.

输入：

First and second line of each input case contain two strings of lowercase character a…z. There are no spaces before, inside or after the strings. Lengths of strings do not exceed 100.

输出：

For each case, output k – the length of a longest common subsequence in one line.

样例输入：

abcd
cxbydz

样例输出：

#include <stdio.h>
#include <string.h>
 
#define MAX 101
 
int main(void){
    char first[MAX], second[MAX];
    int len1, len2;
    int i, j;
    int subseq[2][MAX];
 
    while (scanf ("%s%s", first, second) != EOF){
        len1 = strlen (first);
        len2 = strlen (second);
        for (i=0; i<=len2; ++i)
            subseq[0][i] = 0;
        subseq[1][0] = 0;
        for (i=1; i<=len1; ++i){
            for (j=1; j<=len2; ++j){
                if (first[i-1] == second[j-1])
                    subseq[i%2][j] = subseq[(i-1)%2][j-1] + 1;
                else{
                    subseq[i%2][j] = (subseq[(i-1)%2][j] > subseq[i%2][j-1]) ? subseq[(i-1)%2][j] : subseq[i%2][j-1];
                }
            }
        }
        printf ("%d\n", subseq[len1%2][len2]);
    }
     
    return 0;
}

HDOJ上相似的题目：http://acm.hdu.edu.cn/showproblem.php?pid=1243

参考资料：http://www.cnblogs.com/liyukuneed/archive/2013/05/22/3090597.html