思路：求長度最少的串的思想和求最長公共子串基本一致，dp[i][j]即可。

求數量，則藉助於前面的dp[i][j]。

具體思路看程式碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 40;
char str1[N],str2[N];
int dp[N][N],num[N][N];
int main() 
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int T ,cas = 0;
    scanf("%d", &T); getchar();
    while(T--) {
        gets(str1);
        gets(str2);
        int len1 = strlen(str1), len2 = strlen(str2);

        memset(dp,0,sizeof(dp));
        memset(num,0,sizeof(num));

        for(int i = 1; i < N; i++) 
        {
            dp[i][0] = dp[0][i] = i;
            num[i][0] = num[0][i] = 1;
        }
        num[0][0]=1;
        for(int i=0;i<len1;i++)
        {
            for(int j=0;j<len2;j++)
            {
                if(str1[i]==str2[j])
                {
                    dp[i+1][j+1]=dp[i][j]+1;
                    num[i+1][j+1]=num[i][j];
                }
                else
                {
                    dp[i+1][j+1]=min(dp[i+1][j],dp[i][j+1])+1;
                    if(dp[i+1][j]>dp[i][j+1])
                    {
                        num[i+1][j+1]=num[i][j+1];
                    }
                    else if(dp[i][j+1]>dp[i+1][j])
                    {
                        num[i+1][j+1]=num[i+1][j];
                    }
                    else
                    {
                        num[i+1][j+1]=num[i+1][j]+num[i][j+1];
                    }
                }
            }
        }
        printf("Case #%d: %d %d\n",++cas,dp[len1][len2],num[len1][len2]);
    }
    return 0;
}