Zipper - HDU 1501 dp

最新推荐文章于 2019-04-07 13:47:00 发布

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本文介绍了一个算法问题，探讨如何判断第三字符串是否能由前两个字符串的字符按原顺序组合而成。通过动态规划的方法解决了这一问题，并给出了完整的实现代码。

Zipper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6474 Accepted Submission(s): 2331

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

思路：dp[i][j]第一个字符串的前i个和第二个字符串的前j个能否构成第三个字符串的前i+j个。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
char s[4][310],dp[310][310];
int main()
{ int T,t,i,j,k,len1,len2;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%s%s%s",s[1]+1,s[2]+1,s[3]+1);
    len1=strlen(s[1]+1);
    len2=strlen(s[2]+1);
    if(len1+len2!=strlen(s[3]+1))
    { printf("Data set %d: no\n",t);
      continue;
    }
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(i=0;i<=len1;i++)
     for(j=0;j<=len2;j++)
     { if(i==0 && j==0)
        continue;
       if(s[3][i+j]==s[1][i] && i>0 &&dp[i-1][j]==1)
        dp[i][j]=1;
       if(s[3][i+j]==s[2][j] && j>0 &&dp[i][j-1]==1)
        dp[i][j]=1;
     }
    if(dp[len1][len2]==1)
     printf("Data set %d: yes\n",t);
    else
     printf("Data set %d: no\n",t);
  }
}