hdu 1501 Zipper

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1501

题目描述：

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

3cat tree tcraetecat tree catrteecat tree cttaree

Sample Output

Data set 1: yesData set 2: yesData set 3: no

题目大意：

给你三个字符串，在不改变字符串一和二本身的顺序下，问，能不能组成第三个字符串

题目分析：

DP解法：

最优子结构分析：如上例，如果A、B可以组成C，那么，C最后一个字母e，必定是 A 或 C 的最后一个字母组成。

C去除除最后一位，就变成是否可以求出 A-1和B 或者 A与B-1 与 是否可以构成 C-1。。。

状态转移方程： 用f[i][j] 表示 表示A前 i 为 和B 前j 位是否可以组成 C的前i+j位　　　　　　　　

　　　　　　　　dp[i][j]= (dp[i-1][j]&&(a[i]==c[i+j]))||(dp[i][j-1]&&(b[j]==c[i+j]))

AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int T,dp[205][205],num=1;char s1[205],s2[205],s3[205];int l1,l2,l3;int main(){    cin>>T;    while(T--)    {        scanf("%s%s%s",s1+1,s2+1,s3+1);        l1=strlen(s1+1);        l2=strlen(s2+1);        for(int i=1;i<=l1;i++)        {            if(s1[i]==s3[i])                dp[i][0]=1;        }        for(int i=1;i<=l2;i++)        {            if(s2[i]==s3[i])                dp[0][i]=1;        }        for(int i=1;i<=l1;i++)        {            for(int j=1;j<=l2;j++)            {                dp[i][j]=((dp[i-1][j]&&s1[i]==s3[i+j])||(dp[i][j-1]&&s2[j]==s3[i+j]));            }        }        printf("Data set %d: ",num++);        if (dp[l1][l2]==1)            printf("yes\n");        else            printf("no\n");    }    return 0;}

转自：http://blog.csdn.net/vsooda/article/details/7936229