标签：
Zipper
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7223 Accepted Submission(s): 2576
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
Recommend
linle | We have carefully selected several similar problems for you:
1502 1227 1158 1080 1078
ac代码
#include<stdio.h>
#include<string.h>
char a[220],b[220],c[220*2];
bool flag;
int len1,len2,len3;
void dfs(int x,int y,int len)
{
if(flag)
return;
if(len>=len3)
{
flag=true;
return;
}
if(a[x]==c[len])
{
dfs(x+1,y,len+1);
}
if(b[y]==c[len])
dfs(x,y+1,len+1);
}
int main()
{
int t,cot=0;
scanf("%d",&t);
while(t--)
{
scanf("%s%s%s",a,b,c);
len1=strlen(a);
len2=strlen(b);
len3=strlen(c);
if(len1+len2!=len3)
{
printf("Data set %d: no\n",++cot);
continue;
}
flag=false;
if(a[len1-1]==c[len3-1]||b[len2-1]==c[len3-1])
{
dfs(0,0,0);
}
if(flag)
printf("Data set %d: yes\n",++cot);
else
printf("Data set %d: no\n",++cot);
}
}
HDOJ 题目1501 Zipper（DFS）
标签：
原文地址：http://blog.csdn.net/yu_ch_sh/article/details/44681603
