POJ 2192 Zipper

http://poj.org/problem?id=2192

题意：

给出3个字符串，判断第1个和第2个能否组成第3个，要求是每个字符串的相对顺序不能改变。

 

思路：
d[i][j]表示第1个取第i个字符，第2个取第j个字符时能否组成第3个字符。

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 const int maxn = 500 + 5;
 9 
10 char s1[maxn], s2[maxn], s3[maxn];
11 int d[maxn][maxn];
12 
13 int main()
14 {
15     //freopen("D:\\txt.txt", "r", stdin);
16     int T;
17     scanf("%d", &T);
18     int kase = 0;
19     while (T--)
20     {
21         scanf("%s%s%s", s1+1, s2+1, s3+1);
22         int len1 = strlen(s1+1);
23         int len2 = strlen(s2+1);
24         int len3 = strlen(s3+1);
25         memset(d, 0, sizeof(d));
26         for (int i = 1; i <= len1; i++)
27         {
28             if (s1[i] == s3[i])
29                 d[i][0] = 1;
30             else break;
31         }
32         for (int i = 1; i <= len2; i++)
33         {
34             if (s2[i] == s3[i])
35                 d[0][i] = 1;
36             else break;
37         }
38         for (int i = 1; i <= len1; i++)
39         for (int j = 1; j <= len2; j++)
40         {
41             if (s1[i] == s3[i + j] && d[i - 1][j] == 1)
42                 d[i][j] = 1;
43             if (s2[j] == s3[i + j] && d[i][j - 1] == 1)
44                 d[i][j] = 1;
45         }
46         printf("Data set %d: ", ++kase);
47         if (d[len1][len2] == 1)
48             printf("yes\n");
49         else
50             printf("no\n");
51     }
52 }

 

posted @ 2017-03-17 21:26  Kayden_Cheung  阅读(123)  评论(0)  编辑  收藏  举报
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