題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1503
題意：給了兩個字串，找出它們的最長公共子序列和剩下的兩部分拼接後的串
分析：利用LCS過程，標誌出兩個串不同長度時的狀態，再回溯，逆向輸出結果
程式碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
const int maxn = 110;
typedef long long ll;
char s1[maxn],s2[maxn];
int dp[maxn][maxn],mark[maxn][maxn];

void work(int len1,int len2)
{
    for(int i = 0; i <= len1; i++)
        mark[i][0] = 1;
    for(int i = 0; i <= len2; i++)
        mark[0][i] = -1;
    for(int i = 1; i <= len1; i++)
    {
        for(int j = 1; j <= len2; j++)
        {
            if(s1[i - 1] == s2[j - 1])
            {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                mark[i][j] = 0;
            }
            else if(dp[i][j - 1] >= dp[i - 1][j])
            {
                dp[i][j] = dp[i][j - 1];
                mark[i][j] = -1;
            }
            else
            {
                dp[i][j] = dp[i - 1][j];
                mark[i][j] = 1;
            }
        }

    }
}
void printlcs(int len1,int len2)
{
    //printf("1\n");
    if(!len1 && !len2)
        return;
    if(mark[len1][len2] == 0)
    {
        printlcs(len1 - 1,len2 - 1);
        printf("%c",s1[len1 - 1]);
    }
    else if(mark[len1][len2] == -1)
    {
        printlcs(len1,len2 - 1);
        printf("%c",s2[len2 - 1]);
    }
    else
    {
        printlcs(len1 - 1,len2);
        printf("%c",s1[len1 - 1]);
    }
}
int main()
{
    int n,len1,len2;
    while(~scanf("%s%s",s1,s2))
    {
        memset(dp,0,sizeof(dp));
        len1 = strlen(s1);
        len2 = strlen(s2);
        work(len1,len2);
        printlcs(len1,len2);
        printf("\n");
    }
    return 0;
}