*all in all <poj 1936>
View Code 
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 char a[100005],b[100005];
 6 int ans[2][100005];
 7 void solve()
 8 {
 9     int len1 = strlen(a);
10     int len2 = strlen(b);
11     memset(ans,0,sizeof(ans));
12     for(int i=0;i<len1;i++)
13     {
14         for(int j=0;j<len2;j++)
15         {
16             if((i+1) & 1){
17                 if(a[i] == b[j])
18                 ans[1][j+1] = ans[0][j] + 1;
19                 else
20                 ans[1][j+1] = max(ans[0][j+1],ans[1][j]);
21             }
22             else
23             {
24                if(a[i] == b[j])
25                 ans[0][j+1] = ans[1][j] + 1;
26                 else
27                 ans[0][j+1] = max(ans[1][j+1],ans[0][j]);
28             }
29         }
30     }
31     if(ans[len1&1][len2] == len1 ) cout << "Yes" << endl;
32     else cout << "No" << endl;
33 }
34 int main()
35 {
36     while(cin >> a >> b)
37     {
38         solve();
39     }
40     return 0;
41 }
 posted on 2011-04-26 14:55  eth0  阅读(131)  评论(0)  编辑  收藏  举报