9度OJ 1042 Coincidence--ダイナミック企画（最長共通サブシーケンス）

タイトルの住所:http://ac.jobdu.com/problem.php?pid=1042
テーマの説明:
Find a longest common subsequence of two stings.
入力:
First and second line of each input case contain two streings of lowercase character a...z.The areのspaces before、inside or after the streings.Lengths of stregs do exceed 100.
出力:
For each case，output k–the length of a longest common subsequence in one line.
サンプル入力：

abcd
cxbydz

サンプル出力:

2

#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
", subseq[len1%2][len2]);
    }
     
    return 0;
}

HDOJで似たようなテーマ：http://acm.hdu.edu.cn/showproblem.php?pid=1243
参考資料：http://www.cnblogs.com/liyukuneed/archive/2013/05/22/3090597.html