hdu1501 Zipper DFS带剪枝

最新推荐文章于 2021-10-21 18:26:48 发布

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本文探讨了如何通过组合两个字符串的字符来形成第三个字符串，并详细介绍了实现这一过程的算法。

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Zipper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3613 Accepted Submission(s): 1296

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

注意剪纸不剪纸超时

Source

＃include<stdio.h>
＃include<stdlib.h>
＃include<string.h>
char num1[224]，num2[224]，num3[424];
int len1，len2，len3，flag，hash[224][224];
void DFS（ int a，int b，int c ）
{
     if（ len3==c ）  
     {
         flag=1;
         return ;
     }
     if（ hash[a][b]==1 ） //若是该点已经标识表记标帜就代表该点以下的所有路线都接见过，就不须要持续深搜。

       return ;
     hash[a][b]=1;
     if（ num1[a]==num3[c] ）
     {
           DFS（ a+1，b，c+1 ）;
     }
     if（ num2[b]==num3[c] ）
     {
         DFS（ a，b+1，c+1 ）;    
     }    
}
int main（）
{
    int T;
    scanf（ "％d"，&T ）;
    for（ int i=1; i<=T; i++ ）
    {
          flag=0;
          scanf（ "％s％s％s"，num1，num2，num3 ）;
          len1=strlen（  num1 ），len2=strlen（ num2 ），len3=strlen（ num3 ）;
          memset（ hash，0，sizeof（ hash ） ）;
          DFS（ 0，0，0 ）;
          printf（flag== 1？ "Data set ％d: yes\n": "Data set ％d: no\n"， i）;     
    } 
    return 0;  
}