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最新推荐文章于 2020-07-12 21:05:55 发布

ych19871020

最新推荐文章于 2020-07-12 21:05:55 发布

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本文介绍了一个字符串问题：如何判断第三个字符串是否能由前两个字符串按原顺序交替组合而成。通过一个递归深度优先搜索算法实现，并提供了完整的C++代码示例。

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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

#include<iostream>
using namespace std;
bool dp[201][201];//记忆化搜索，存储该状态是否检索过
char fir[201],sec[201],thi[402];
int n,firLen,secLen,thiLen;
bool isOk;
void dfs(int i,int j,int k){
    if(k==thiLen){
        isOk=true;
        return;
    }
    if(isOk)return ;//已经找到，直接跳出
    
    if(dp[i][j])return;//该状态搜索过直接跳出
    
    if(i<firLen&&fir[i]==thi[k])
        dfs(i+1,j,k+1);
    if(j<secLen&&sec[j]==thi[k])
        dfs(i,j+1,k+1);
    
    dp[i][j]=true;//做上标记，表明此状态扩展开的dfs都搜过了
    
}

int main(){
    while(scanf("%d",&n)!=EOF){
    for(int i=1;i<=n;i++){
        isOk=false;
        memset(dp,false,sizeof(dp));
        scanf("%s%s%s",fir,sec,thi);

        firLen=strlen(fir);
        secLen=strlen(sec);
        thiLen=strlen(thi);

        dfs(0,0,0);
        cout<<"Data set "<<i<<": ";
        cout<<(isOk?"yes":"no")<<endl;
    }
    }
    return 0;
}