Poj 1159 palindrome (string-to-text: LCs)

Poj 1159 palindrome (string-to-text: LCs)

Http://poj.org/problem? Id = 1159

Question:

I will give you a string and ask you if you need to insert a few characters into the string, but it will become a return string.

Analysis:

First, match the original string with its inverse string to find the longest common subsequence. the string of the longest common subsequence must be a return string. therefore, the rest of the original string does not constitute a background. we only need to add the remaining part of the characters to the corresponding position, and the original string naturally becomes a background.

So the answer to this question is: N minus (the LCS length of the original string and the inverse string ).

For example, DP [I] [J] = X indicates the longest common subsequence of the first I character of string a and the first J character of string B.

Initialization: DP is all 0.

Status transfer:

A [I] = B [J]: DP [I] [J] = DP [I-1] [J-1] + 1.

A [I]! = B [J]: DP [I] [J] = max (DP [I-1] [J], DP [I] [J-1]).

Final requirement: DP [N] [M].

The two-dimensional rolling array used by the program. If int [5000] [5000] is used, the memory will be exceeded.

AC code:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=5000+5;int n;char s1[maxn],s2[maxn];int dp[2][maxn];int main(){    while(scanf("%d",&n)==1)    {        scanf("%s",s1);        for(int i=0;i<n;i++)            s2[i]=s1[n-1-i];        memset(dp,0,sizeof(dp));        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-1)%2][j] , dp[i%2][j-1]);        }        printf("%d\n",n-dp[n%2][n]);    }    return 0;}

Poj 1159 palindrome (string-to-text: LCs)