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