HDU - 1501 Zipper（dp&深搜）

题目：HDU - 1501 Zipper

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

题意理解：

• 给你三个字符串，问你前两个是否能组成第三个串，串中字符顺序不可改变，可拆分。

思路(dp)：

简单的dp写法，用一个二维数组，第一维代表第一个字符串用了前几位，第二维代表第二个字符串用了前几位，若 dp[i][j] 可以组成 str[i+j]之前的所有字符，则为1，否为0。 例子： abc def adebcf dp为1的状态都有： dp[1][0] = 1; 取str1[0] dp[1][1] = 1; 在dp[1][0]的基础上再取str2[0],以此类推 dp[1][2] = 1; dp[2][2] = 1; dp[3][2] = 1; dp[3][3] = 1;

代码行(dp)：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<deque>
#include<cstdlib>
#include<fstream>
#include<map>
#include<iomanip>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<list>
typedef long long ll;
using namespace std;
string str1,str2,str;
int dp[440][440];
	
int main()
{

	int T;
	scanf("%d", &T);
	for(int k=0;k<T;k++)
	{
		cin>>str1>>str2>>str;

		int l1=str1.length(),
		    l2=str2.length(),
		    l=str.length();
		printf("Data set %d: ", k+1);
		if(l!=l1+l2)
		{
			cout<<"no"<<endl;
			continue;
		}
		memset(dp,0,sizeof(dp));
		dp[0][0]=1;
		for(int i=0;i<=l1;i++)
		{
			for(int j=0;j<=l2;j++)
			{
				if(i==0 && j==0) continue;
				if(i!=0 && dp[i-1][j]==1 && str1[i-1]==str[i+j-1])
				{
					dp[i][j]=1;
					continue;
				}
				if(j!=0 && dp[i][j-1]==1 && str2[j-1]==str[i+j-1])
				{
					dp[i][j]=1;
				}
			}
			
		}
		if(dp[l1][l2]==1) cout<<"yes"<<endl;
			else cout<<"no"<<endl;
	}
	return 0;
}