HUD 1501 Zipper（记忆化 or DP)

Problem Description

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat

String B: tree

String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat

String B: tree

String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

Input

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two
strings will have lengths between 1 and 200 characters, inclusive.

Output

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

Sample Input

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

Sample Output

Data set 1: yes
Data set 2: yes
Data set 3: no

Source

Pacific Northwest 2004

记忆化和DP均可以，通过两个串合成第三个串。顺序不变。

记忆化：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
const int maxn=220;
char s1[maxn],s2[maxn],s[maxn<<1];
int dp[maxn][maxn];
int len,len1,len2;
int t;
int dfs(int i,int j,int k)
{
   if(k==len)
      return 1;
   if(dp[i][j])
        return 0;
   dp[i][j]=1;
   if(s1[i]==s[k])
   {
       if(dfs(i+1,j,k+1))
        return 1;
   }
   if(s2[j]==s[k])
   {
       if(dfs(i,j+1,k+1))
        return 1;
   }
   return 0;
}
int main()
{
    int l=0;
    scanf("%d",&t);
    while(t--)
    {
         scanf("%s%s%s",s1,s2,s);
         len1=strlen(s1);
         len2=strlen(s2);
         len=strlen(s);
         memset(dp,0,sizeof(dp));
         if(dfs(0,0,0))
            printf("Data set %d: yes\n",++l);
         else
            printf("Data set %d: no\n",++l);
    }
    return 0;
}

DP:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=220;
char s1[maxn],s2[maxn],s[maxn<<1];
int dp[maxn][maxn];

int main()
{
    int t;
    int l=0;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        scanf("%s%s%s",s1+1,s2+1,s+1);
        int len1=strlen(s1+1);
        int len2=strlen(s2+1);
        int len=strlen(s+1);
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=0;i<=len1;i++)
        {
           for(int j=0;j<=len2;j++)
           {
               if(i==0&&j==0)
                  continue;
               if(s[i+j]==s1[i]&&i&&dp[i-1][j])
                  dp[i][j]=1;
               if(s[i+j]==s2[j]&&j&&dp[i][j-1])
                  dp[i][j]=1;
           }
        }
        if(dp[len1][len2])
           printf("Data set %d: yes\n",++l);
        else
           printf("Data set %d: no\n",++l);
    }
    return 0;
}

HUD 1501 Zipper（记忆化 or DP)