LightOJ 1110 - An Easy LCS(最长公共子序列 路径输出 三维 模板啊)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1110

LCS means 'Longest Common Subsequence' that means two non-empty strings are given; you have to find the Longest Common Subsequence between them. Since there can be many solutions, you have to print the one which is the lexicographically smallest. Lexicographical order means dictionary order. For example, 'abc' comes before 'abd' but 'aaz' comes before 'abc'.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a blank line. The next two lines will contain two strings of length 1 to 100. The strings contain lowercase English characters only.

Output

For each case, print the case number and the lexicographically smallest LCS. If the LCS length is 0 then just print ':('.

Sample Input

Output for Sample Input

3

 

ab

ba

 

zxcvbn

hjgasbznxbzmx

 

you

kjhs

Case 1: a

Case 2: zxb

Case 3: :(

PS:

字典序输出！

转自：http://blog.sina.com.cn/s/blog_7fec19cd010112jm.html

代码如下：

/*最长公共子序列:dp[][]存的是到当前位置，最长序列的个数，s[][][]存的是到当前位置最长的公共子序列，其实dp[][]，和s[][][]的原理是一样的，不同的是，s[][]存的是一个序列，所以是三维的*/#include <stdio.h>#include <string.h>#define N 150int dp[N][N];char s[N][N][N];char s1[N], s2[N];int main(){    int t,i,j,k;    int cas = 0;    scanf("%d",&t);    while(t--)    {        scanf("%s%s",s1+1,s2+1);        memset(s,0,sizeof(s));        for(i = 1; s1[i]; i++)            for(j = 1; s2[j]; j++)            {                if(s1[i] == s2[j])                {                    dp[i][j] = dp[i-1][j-1]+1;                    for(k = 0; k < dp[i-1][j-1]; k++)                        s[i][j][k] = s[i-1][j-1][k];                    s[i][j][k] = s1[i];                }                else                {                    if(dp[i-1][j] > dp[i][j-1])                    {                        dp[i][j] = dp[i-1][j];                        strcpy(s[i][j],s[i-1][j]);                    }                    else if(dp[i][j-1] > dp[i-1][j])                    {                        dp[i][j] = dp[i][j-1];                        strcpy(s[i][j],s[i][j-1]);                    }                    else                    {                        dp[i][j] = dp[i-1][j];                        if(strcmp(s[i-1][j],s[i][j-1])>0)                            strcpy(s[i][j],s[i][j-1]);                        else                            strcpy(s[i][j],s[i-1][j]);                    }                }            }        printf("Case %d: ",++cas);        i--;        j--;        if(dp[i][j] == 0)//长度            printf(":(\n");        else            printf("%s\n",s[i][j]);    }    return 0;}