Palindrome

http://poj.org/problem?id=1159

最少需要补充的字母数=x的长度-x和y的最长公共子序列的长度。

状态的转移方程：

if(i==0||y==0) dp[i][j]=0;

else if(x[i]==y[j]) dp[i][j]=dp[][i-1][j-1]+1;

else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 int a[2][5005];
 9 int main()
10 {
11     int n;
12     string s1,s2;
13     while(scanf("%d",&n)!=EOF){
14 
15         cin>>s1;
16         s2=s1;
17         reverse(s1.begin(),s1.end());
18         memset(a,0,sizeof(a));
19         for(int i=1;i<=n;i++)
20         {
21             for(int j=1;j<=n;j++)
22             {
23                 if(s1[i-1]==s2[j-1])
24                 {
25                     a[i%2][j]=a[(i-1)%2][j-1]+1;
26                 }
27                 else
28                 {
29                     a[i%2][j]=max(a[(i-1)%2][j],a[i%2][j-1]);
30                 }
31             }
32         }
33         printf("%d\n",n-a[n%2][n]);
34     }
35     return 0;
36 }
View Code 
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 int a[2][5005];
 9 int main()
10 {
11     int n;
12     string s1,s2;
13     while(scanf("%d",&n)!=EOF){
14 
15         cin>>s1;
16         s2=s1;
17         reverse(s1.begin(),s1.end());
18         memset(a,0,sizeof(a));
19         for(int i=1;i<=n;i++)
20         {
21             for(int j=1;j<=n;j++)
22             {
23                 a[i%2][j]=max(a[(i-1)%2][j],a[i%2][j-1]);
24                 if(s1[i-1]==s2[j-1])
25                 {
26                     a[i%2][j]=max(a[i%2][j],(a[(i-1)%2][j-1]+1));
27                 }
28             }
29         }
30         printf("%d\n",n-a[n%2][n]);
31     }
32     return 0;
33 }
View Code

 

posted @ 2013-08-13 23:14  null1019  阅读(222)  评论(0)  编辑  收藏  举报