【OpenJ_Bailian - 2192】Zipper

Zipper

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Descriptions:

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

 题目链接：

https://vjudge.net/problem/OpenJ_Bailian-2192

题意：给定三个字符串，A，B，C。A和B的长度和等于C。判断字符串C能否由字符串A、B中的字符组成。要求，原来A 和 B 中的字符作为C中的字符时，还必须保持字符在原串中的顺序。

假设 A中的前 i 个字符 和  B 中的前 j 个字符 可以构成 C中的前 i+j个字符。那么 如果A中的前 i+1 个字符和B中的前 j 个字符可以构成 C 中的前 i+j+1 个字符 或者 如果A中的前 i 个字符和B中的前 j+1 个字符可以构成 C 中的前 i+j+1 个字符。那么显然 C 中的前 i+j+1 个字符可由字符串A 和 字符串 B 中的前 i+j+1 个字符构成。没啥说的直接上代码吧

AC代码

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 using namespace std;
15 string s1,s2,s3;
16 int cnt1,cnt2,cnt3;
17 int len1,len2,len3;
18 bool dp[1005][1005];
19 bool f=0;//建一个flag，判断输出yes/no
20 void dfs(int t1,int t2,int t3)//深搜
21 {
22     if(f)
23         return;
24     if(t1==len1&&t2==len2)
25     {
26         f=1;
27         return;
28     }
29     if(dp[t1][t2]==1)
30         return;
31     if(s1[t1]==s3[t3])
32     {
33         dp[t1][t2]=1;
34         dfs(t1+1,t2,t3+1);
35     }
36     if(s2[t2]==s3[t3])
37     {
38         dp[t1][t2]=1;
39         dfs(t1,t2+1,t3+1);
40     }
41 }
42 int main()
43 {
44     int sum=0;
45     int T;
46     cin >> T;
47     while(T--)
48     {
49         f=0;
50         sum++;
51         memset(dp,0,sizeof(dp));//必须初始化
52         cin >> s1 >> s2 >> s3;
53         len1=s1.length();
54         len2=s2.length();
55         len3=s3.length();
56         dfs(0,0,0);
57         if(f)
58             printf("Data set %d: yes\n",sum);
59         else
60             printf("Data set %d: no\n",sum);
61     }
62 }