杭电-1501Zipper（dfs+标记数组）

向右看齐 2022-08-17 15:14 95阅读 0赞

# Zipper #

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

Total Submission(s): 8428 Accepted Submission(s): 2976

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

用dfs每次把总数组分给两个数组，但是一定要用个数组标记每种状态，否则会超时！

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int flag;
    int len1,len2,len,vis[300][600];
    char str1[300],str2[300],str[600];
    void dfs(int a,int b,int c)
    {
        if(a==len1&&b==len2&&c==len)
        {
            flag=1;
            return ;
        }
        if(str[c]!=str1[a]&&str[c]!=str2[b])
            return ;
        if(a>len1||b>len2||c>len)
            return ;
            if(vis[b][c])
                return ;
            vis[b][c]=1;
        if(flag)
            return ;
        if(str[c]==str1[a])
            dfs(a+1,b,c+1);
    
        if(str[c]==str2[b])
            dfs(a,b+1,c+1);
    }
    int main()
    {
        int N,cot=1;
    
        scanf("%d",&N);
        while(N--)
        {
            scanf("%s%s%s",str1,str2,str);
            memset(vis,0,sizeof(vis));
            len1=strlen(str1);
            len2=strlen(str2);
            len=strlen(str);
            flag=0;
            dfs(0,0,0);
            if(flag)
                printf("Data set %d: yes\n",cot++);
            else
                printf("Data set %d: no\n",cot++);
    
        }
    }