Light oj 1110 - An Easy LCS（LCS）

1110 - An Easy LCS

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Time Limit: 2 second(s)Memory Limit: 32 MB

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: :(

 

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PROBLEM SETTER: JANE ALAM JAN

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define bug printf("hihi\n")#define eps 1e-12typedef __int64 ll;using namespace std;#define N 105int dp[N][N];char f[N][N][N];char a[N],b[N];int lena,lenb;void DP(){    int i,j;    memset(dp,0,sizeof(dp));    for(i=1;i<=lena;i++)        for(j=1;j<=lenb;j++)    {        if(a[i]==b[j])        {            dp[i][j]=dp[i-1][j-1]+1;            int k;           strcpy(f[i][j],f[i-1][j-1]);            f[i][j][dp[i-1][j-1]]=a[i];        }        else            if(dp[i-1][j]>dp[i][j-1])        {             dp[i][j]=dp[i-1][j];             strcpy(f[i][j],f[i-1][j]);        }        else            if(dp[i-1][j]<dp[i][j-1])        {            dp[i][j]=dp[i][j-1];            strcpy(f[i][j],f[i][j-1]);        }        else           if(dp[i-1][j]==dp[i][j-1])           {               dp[i][j]=dp[i-1][j];                if(strcmp(f[i-1][j],f[i][j-1])<0)                      strcpy(f[i][j],f[i-1][j]);                else                    strcpy(f[i][j],f[i][j-1]);           }    }}int main(){    int i,j,t,ca=0;    scanf("%d",&t);    while(t--)    {        scanf("%s%s",a+1,b+1);        lena=strlen(a+1);        lenb=strlen(b+1);        memset(f,0,sizeof(f));        DP();        printf("Case %d: ",++ca);        if(dp[lena][lenb]==0)            printf(":(\n");        else            printf("%s\n",f[lena][lenb]);    }    return 0;}