【hdu1501】zipper——dfs

题目：

Zipper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8645 Accepted Submission(s): 3061



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

Pacific Northwest 2004

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描述：给两个字符串，问能否归并成第三个字符串

题解：用一个二维数组来记录是否枚举过某个情况。由于存在a，b两个数组的某一个元素都可以去放在第三个数组相同值元素的位置，所以不能用队列，要dfs记忆化搜索，分三种情况，两个都行的，其中一个行一个不行的。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
char a[500], b[500], c[500];
int vis[220][220];
bool ans = false;
void init()
{
memset(vis, 0, sizeof(vis));
ans = false;
}

void dfs(int i, int j, int k)
{
if (vis[i][j])
return;
vis[i][j] = 1;
if (c[k] == '\0')
{
ans = true;
return;
}
if (a[i] == c[k] && b[j] == c[k])
{
dfs(i + 1, j, k + 1);
//vis[i + 1][j] = 0;                   //由于是点对来标记，其实不需要回溯，因为这个点不会再被跑到了
dfs(i, j + 1, k + 1);
}
else if (a[i] == c[k] && b[j] != c[k])
dfs(i + 1, j, k + 1);
else if (a[i] != c[k] && b[j] == c[k])
dfs(i, j + 1, k + 1);
else
return;
}

int main()
{
//freopen("input.txt", "r", stdin);
int T, cnt = 1;
cin >> T;
while (T--)
{
scanf("%s%s%s", a, b, c);

init();
dfs(0, 0, 0);
if (ans)
printf("Data set %d: yes\n", cnt++);
else
printf("Data set %d: no\n", cnt++);
}
return 0;
}