poj1159 Palindrome 线性dp+滚动数组

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int maxn=5005;
int n;
string s1,s2;
int dp[2][maxn];//dp[i][j]只依赖于dp[i-1][j],dp[i][j-1]和dp[i-1][j-1]可用滚动数组节约空间,在时间上没有优势
int main(){
	while(cin>>n){
		memset(dp,0,sizeof(dp)); 
		cin>>s1;
		s2=s1;
		reverse(s2.begin(),s2.end());//反转字符串
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(s1[i-1]==s2[j-1]){
					dp[i%2][j]=dp[(i-1)%2][j-1]+1;
				}
				else{
					dp[i%2][j]=max(dp[i%2][j-1],dp[(i-1)%2][j]);
				}
			}
		}
		cout<<n-dp[n%2][n]<<endl;
	}
	return 0;
}

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转载自blog.csdn.net/skykone1/article/details/88971265