Strings - UVa 11081 dp

最新推荐文章于 2020-02-04 20:59:46 发布

提比-我有特殊的AC技巧

最新推荐文章于 2020-02-04 20:59:46 发布

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分类专栏： UVa dp动态规划和递推

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本文介绍了一个编程问题，即如何计算由两个字符串的子序列构成第三个特定字符串的不同方式的数量，并提供了一段AC代码作为解决方案。

Problem H
Strings
Input: Standard Input

Output: Standard Output

Given 3 strings of only lowercase letter you have to count the number of ways you can construct the third string by combining two subsequences from the first two strings.

After deleting 0 or more characters from a string we can get its subsequence. For example “a”, “b”, “c”, “ab”, “ac”, “bc” and “abc” all the strings are the subsequences of “abc”. A subsequence may also be empty.

Now suppose there are two subsequences “abc” and “de”. By combining them you can get the following strings “abcde”, “abdce”, “abdec”, “adbce”, “adbec”, “adebc”, “dabce”, “dabec”, “daebc” and “deabc”.

Input

The first line of the input contains a single integer T (0<T<271) indicating the number of test cases. Each test case contains 3 strings containing only lowercase characters. The lengths of the strings are between 1 and 60.

Output

For each test case output a single integer denoting the number of ways you can construct the third string from the first two string by the above way. The result may be very large. You should output the result%10007.

Sample Input Output for Sample Input

abc abc abc

abbcd bccde abcde

题意：由前两个字符串的子串组成第三个字符串，问有多少种方法。

思路：f1[i][j][k]表示第k个是从第一个字符串取的方法，f2[i][j][k]表示第k个是从第二个字符串取的方法，f表示f1和f2对应的和，然后递推。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s1[110],s2[110],s3[110];
int f1[110][110][110],f2[110][110][110],f[110][110][110];
int main()
{ int t,i,j,k,len1,len2,len3;
  scanf("%d",&t);
  while(t--)
  { scanf("%s%s%s",s1+1,s2+1,s3+1);
    len1=strlen(s1+1);
    len2=strlen(s2+1);
    len3=strlen(s3+1);
    memset(f1,0,sizeof(f1));
    memset(f2,0,sizeof(f2));
    memset(f,0,sizeof(f));
    for(i=0;i<=len1;i++)
     for(j=0;j<=len2;j++)
     { f1[i][j][0]=1;
       f2[i][j][0]=1;
       f[i][j][0]=1;
     }
    for(k=1;k<=len3;k++)
     for(i=0;i<=len1;i++)
      for(j=0;j<=len2;j++)
      { if(i==0 && j==0)
         continue;
        if(i>0)
        { f1[i][j][k]=f1[i-1][j][k];
          if(s1[i]==s3[k])
           f1[i][j][k]+=f[i-1][j][k-1];
        }
        if(j>0)
        { f2[i][j][k]=f2[i][j-1][k];
          if(s2[j]==s3[k])
           f2[i][j][k]+=f[i][j-1][k-1];
        }
        f[i][j][k]=(f1[i][j][k]+f2[i][j][k])%10007;
      }
    printf("%d\n",f[len1][len2][len3]);
  }
}