HDOJ 1501 Zipper（dfs）

Zipper

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7781    Accepted Submission(s): 2754

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
  
  

 

 

 

 

经过这题，对搜索中的递归回溯终于看清了点眉目。看搜索代码最让自己崩溃的就是递归函数。之前一直认为递归的返回值直接返回到主函数，结束递归。今天调试这题的递归函数代码，发现递归中的返回值大部分情况是返回到上次调用函数的地方的真假值判断。

递归是一个回溯过程，不断朝目标接近，再将目标一步步拉出来，不断通过函数中的语句运算，得出题目所要求的值。（不知道这么说对不对，就算对了估计别人也看不懂来年的学弟学妹，看到了不要笑话我，本渣尽力了）

 

 

 

题目大意：给出三个字符串，第三个由前两个自由组合成的，但不能改变字母在原字符串中的前后位置。写一个程序判断给出的三个字符串是否满足上述要求。

 

题目不算难，至少比迷宫好多了，代码如下：

 

 

#include<stdio.h>
#include<string.h>
char a[210],b[210],c[410];
int v[210][210],lena,lenb,lenc;
int dfs(int i,int j,int k)
{
	if(c[k]=='\0')
	   return 1;
	if(v[i][j])//每个位置只能用一次，不进行这步剪枝会超时 
	   return 0;
	v[i][j]=1;
	if(a[i]==c[k]&&dfs(i+1,j,k+1))
	    return 1;
	if(b[j]==c[k]&&dfs(i,j+1,k+1))
	    return 1;
	return 0;//a，b中的字母都找不到c[k]就结束查找 
} 

int main()
{
	int t,n=1;
	scanf("%d",&t);
	while(t--)
	{
		memset(v,0,sizeof(v));
		scanf("%s%s%s",a,b,c);
		lena=strlen(a);
		lenb=strlen(b);
		lenc=strlen(c);
		printf("Data set %d: ",n);
		if(dfs(0,0,0))
		    printf("yes\n");
		else
		    printf("no\n");
		n++;
	}
	return 0;
}