标签：poj hdu dfs
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题目链接
HDU：http://acm.hdu.edu.cn/showproblem.php?pid=1501
POJ: http://poj.org/problem?id=2192
Zipper
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
题意：就是找在不变换字符串1和字符串2的字母的顺序的情况下，能否组成字符串3；
代码如下：
#include <cstdio>
#include <cstring>
#define N 247
char a[N], b[N], c[N+N];
int len1, len2, len3;
int dfs(int x, int y, int sum)
{
if(sum >len3)
return 1;
if(a[x]!=c[sum] && b[y]!=c[sum])
return 0;
if(a[x]==c[sum] && dfs(x+1,y,sum+1))
return 1;
if(b[y]==c[sum] && dfs(x,y+1,sum+1))
return 1;
return 0;
}
int main()
{
int t, cas = 0;
while(~scanf("%d",&t))
{
while(t--)
{
scanf("%s%s%s",a+1,b+1,c+1);//从字符串下标为1开始输入
len1 = strlen(a+1);
len2 = strlen(b+1);
len3 = strlen(c+1);
int flag = 0;
if(c[len3]==a[len1] || c[len3]==b[len2])//优化，如果C最后的字母是a的最后一个
{ //或者b的最后一个才有可能是由它们组成的
flag = dfs(1, 1, 1);
}
if(flag == 1)
printf("Data set %d: yes\n",++cas);
else
printf("Data set %d: no\n",++cas);
}
}
return 0;
}
hdu1501&&poj2192 Zipper(DFS),布布扣,bubuko.com
hdu1501&&poj2192 Zipper(DFS)
标签：poj hdu dfs
原文地址：http://blog.csdn.net/u012860063/article/details/36390665
