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]);

#include<cstdio>
#include<iostream>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int a[2][5005];
int main()
{
    int n;
    string s1,s2;
    while(scanf("%d",&n)!=EOF){

        cin>>s1;
        s2=s1;
        reverse(s1.begin(),s1.end());
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(s1[i-1]==s2[j-1])
                {
                    a[i%2][j]=a[(i-1)%2][j-1]+1;
                }
                else
                {
                    a[i%2][j]=max(a[(i-1)%2][j],a[i%2][j-1]);
                }
            }
        }
        printf("%d
",n-a[n%2][n]);
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<cmath>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int a[2][5005];
int main()
{
    int n;
    string s1,s2;
    while(scanf("%d",&n)!=EOF){

        cin>>s1;
        s2=s1;
        reverse(s1.begin(),s1.end());
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                a[i%2][j]=max(a[(i-1)%2][j],a[i%2][j-1]);
                if(s1[i-1]==s2[j-1])
                {
                    a[i%2][j]=max(a[i%2][j],(a[(i-1)%2][j-1]+1));
                }
            }
        }
        printf("%d
",n-a[n%2][n]);
    }
    return 0;
}