HDU1501-Zipper-字符串的dfs

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". 

InputThe 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.

OutputFor 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

题意：
给出a、b、c，判断a和b按照顺序拼接是否能构成c

具体思路见代码注释，实在是感觉深搜一层层递归回去不好解释

 #include<stdio.h>
 #include<iostream>
 #include<string.h>
 #define inf 0x3f3f3f3f
 using namespace std;

 char a[],b[],c[];//注意一下c的长度开的大小得是ab之和
 int flag,la,lb,lc;
 bool book[][];//记得要标记，否则不知道是否用过，会造成重复使用

 void dfs(int x,int y,int z)////传入下标
 {
     if(z==lc)//长度找到之后
     {
         flag=;
         return;
     }
     if(flag)//已经找到了//该字符已经用过了
         return;
     if(book[x][y]==)//该字符已经用过了
         return;
     book[x][y]=;
     if(x<la&&a[x]==c[z])//a的第x个字符跟c串的第z个字符相等
         dfs(x+,y,z+);//加1就说明是按照顺序来的
     if(y<lb&&b[y]==c[z])//b的第y个字符跟c串的第z个字符相等
         dfs(x,y+,z+);
 }

 int main()
 {
     int tt,t=;
     scanf("%d",&tt);
     while(tt--)
     {
         memset(book,,sizeof(book));
 //        memset(a,'\0',sizeof(a));
 //        memset(b,'\0',sizeof(b));
 //        memset(c,'\0',sizeof(c));
         scanf("%s %s %s",a,b,c);
         la=strlen(a);
         lb=strlen(b);
         lc=strlen(c);
         flag=;
         dfs(,,);
         if(flag)
             printf("Data set %d: yes\n",t++);
         else
             printf("Data set %d: no\n",t++);
     }
     return ;
 }