POJ-2192 Zipper

题目

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

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题解

首先明确几点：

• 串C的长度等于串A+串B

• 串C中每个字符要么来自串A要么来自于串B

于是我们可以做出以下推导:

• StrC==NULL—>A==NULL&&B===NULL
• StrC+C[1]—>Len(StrC)+1—>Len(StrA)+1||Len(StrB)+1
• 因为C中的每个字符都是来自于A或者B—>C[1]==A[1]||C[1]==B[1]

因为长度相等，所以C[i]不是来自于A[j]就是来自于B[i−j],所以我们只需表示A[j]即可

于是DP递推式表示如下：

• DP[i][j]:i,StrC的下标,j:StrA的下标
• DP[All][All]=0,DP[0][0]=1
• StrC[i+1]==StrA[i+1]–>DP[i+1][j+1]=1
• StrC[i+1]==StrB[i+1−j]—>DP[i+1][j]=1
• DP[LenC][LenA]==1–>Yes

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代码

/*********************************Author: ToudsourCreated Time: 五  8/14 19:17:24 2015File Name:D.cpp*********************************/#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;char StrA[210];char StrB[210];char StrC[410];int DP[410][210];int main(){    int T;    scanf("%d",&T);    for(int Case=1;Case<=T;Case++)    {        int LenA;        int LenB;        int LenC;        scanf("%s",StrA+1);        scanf("%s",StrB+1);        scanf("%s",StrC+1);         LenA=strlen(StrA+1);        LenB=strlen(StrB+1);        LenC=strlen(StrC+1);        memset(DP,0,sizeof(DP));        DP[0][0]=1;        for(int i=0;i<=LenC;i++)        {            for(int j=0;j<=LenA;j++)                if(DP[i][j]==1)                {                    //cout<<StrT[i+1]<<" "<<StrO[j+1]<<" "<<StrD[i+1-j]<<endl;                    if(StrC[i+1]==StrA[j+1])                        DP[i+1][j+1]=1;                    if(StrC[i+1]==StrB[i+1-j]&&(i+1-j)<=LenD)                        DP[i+1][j]=1;                }        }        if(DP[LenC][LenA])            printf("Data set %d: yes\n",Case);        else            printf("Data set %d: no\n",Case);    }    return 0;}