杭电 1513 Palindrome LCS DP

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3955    Accepted Submission(s): 1349

Problem Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

 

 

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

 

 

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

 

 

Sample Input

5Ab3bd

 

 

Sample Output

2

 

第一次用滚动数组，感觉自己萌萌哒QAQ。

嘛，滚动数组其实并不难，关键是要用i来确定，这样会比较方便点。虽然还是有点小不理解吧，不过这都不是事儿~

#include<stdio.h>#include<string.h>#define maxn 5000+10#define max(a,b) (a>b?a:b)char s1[maxn],s2[maxn];int dp[2][maxn];int main(){int n;while(~scanf("%d",&n)){scanf("%s",s1);strcpy(s2,s1);strrev(s2);memset(dp,0,sizeof(dp));int i,j,x=0,y=1;for(i=1;i<=n;i++){for(j=1;j<=n;j++){x=i%2;y=1-x;if(s1[i-1]==s2[j-1])dp[x][j]=dp[y][j-1]+1;elsedp[x][j]=max(dp[y][j],dp[x][j-1]);}}printf("%d\n",n-dp[n%2][n]);}return 0;}