# poj 2192 Zipper

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18658 Accepted: 6651

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
```

Source

Pacific Northwest 2004

1、  第三个串最后一个字符要么是串1的最后一个字符，要么是串2的最后一个字符

2、  按照串1的顺序对串3进行搜索，若不匹配则该字符必是串2的下一个字符。

``` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 char first[202],second[202],third[402],Left[401];
6 int sign[402];
7 bool flag;
8 int check()
9 {
10     int i,count=0;
11     int k=strlen(third);
12     for(i=0;i<k;i++)
13         if(!sign[i]) Left[count++]=third[i];
14     Left[count]='\0';
15     if(strcmp(Left,second)==0) return 1;
16     return 0;
17 }
18 int dfs(int f,int s,int t)
19 {
20     if(f>=strlen(first))
21     {
22         if(check()) flag=true;
23         return 0;
24     }
25     if(flag) return 0;
26     if(first[f]==third[s])
27     {
28         sign[s]=1;
29         if(s<strlen(third)) dfs(f+1,s+1,t);
30         sign[s]=0;
31     }
32     else
33     {
34         if(third[s]!=second[t]) return 0;//剪枝2
35     }
36     if(!flag && s<strlen(third)) dfs(f,s+1,t+1);
37     return 0;
38 }
39 int main()
40 {
41     int len1,len2,len3,Case,count=0;
42     scanf("%d",&Case);
43     while(Case--)
44     {
45         count++;
46         flag=false;
47         scanf("%s %s %s",first,second,third);
48         memset(sign,0,sizeof(sign));
49
50         len1=strlen(first);
51         len2=strlen(second);
52         len3=strlen(third);
53
54         if(len1+len2!=len3)
55         {
56             printf("Data set %d: no\n",count);
57             continue;
58         }
59         if(third[len3-1]!=first[len1-1] && third[len3-1]!=second[len2-1])// 剪枝1
60         {
61             printf("Data set %d: no\n",count);
62             continue;
63         }
64         dfs(0,0,0);
65         if(flag)
66             printf("Data set %d: yes\n",count);
67         else
68             printf("Data set %d: no\n",count);
69     }
70     return 0;
71 }```
View Code

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：

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

``` 1 #include<stdio.h>
2 #include<string.h>
3
4 char a[201],b[201],c[402];
5 int la,lb,lc;
6 int dp[201][201];
7
8 int main()
9 {
10     int ncase;
11     scanf("%d",&ncase);
12     for(int n=1; n<=ncase; n++) {
13
14         a[0]='p';
15         b[0]='p';
16         c[0]='p';
17
18         scanf("%s%s%s",a+1,b+1,c+1);
19
20         la=strlen(a);
21         lb=strlen(b);
22         lc=strlen(c);
23
24         la-=1;
25         lb-=1;
26
27         //处理边界
28         for (int i=1; i<=la; i++)
29             if (a[i]==c[i]) dp[i][0]=1;
30
31         for (int i=1; i<=lb; i++)
32             if (b[i]==c[i]) dp[0][i]=1;
33         //DP
34         for (int i=1; i<=la; i++)
35             for (int j=1; j<=lb; j++)
36                 dp[i][j]= (dp[i-1][j]&&(a[i]==c[i+j]))||(dp[i][j-1]&&(b[j]==c[i+j]));
37
38         printf("Data set %d: ",n);
39         if (dp[la][lb]==1) printf("yes\n");
40         else printf("no\n");
41
42     }
43 }```

Res[i][j]= (third[i+j]==first[i] && res[i-1][j]==1) ||(third[i+j]==second[j]&&res[i][j-1]==1)

``` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 using namespace std;
5 char first[201],second[201],third[401];
6 int res[201][201];
7 int init(int n,int m)
8 {
9     int i;
10     for(i=1;i<=m;i++)
11         if(second[i]==third[i]) res[0][i]=1;
12         else break;
13     for(i=1;i<=n;i++)
14         if(first[i]==third[i]) res[i][0]=1;
15         else break;
16     return 0;
17 }
18 int dp(int n,int m)
19 {
20     int i,j;
21     for(i=1;i<=n;i++)
22         for(j=1;j<=m;j++)
23         {
24             if(third[i+j]==first[i] && res[i-1][j]) res[i][j]=1;
25             if(third[i+j]==second[j] && res[i][j-1]) res[i][j]=1;
26         }
27     if(res[n][m]) return 1;
28     return 0;
29 }
30 int main()
31 {
32     int n,len1,len2,count=0;;
33     scanf("%d",&n);
34     while(n--)
35     {
36         count++;
37         scanf("%s %s %s",first+1,second+1,third+1);
38         len1=strlen(first+1);
39         len2=strlen(second+1);
40         memset(res,0,sizeof(res));
41         init(len1,len2);
42
43         if(dp(len1,len2))
44             printf("Data set %d: yes\n",count);
45         else
46             printf("Data set %d: no\n",count);
47     }
48     return 0;
49 }```
View Code