POJ 1159 Palindrome（回文串，最长公共子序列）

Palindrome

Time Limit: 3000MS
Memory Limit: 65536KTotal Submissions: 63457
Accepted: 22128

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

Source

IOI 2000

题目大意：给你个字符串，问最少加入多少字符可以让这个字符串变成回文字符串。

解题思路：

最少需要补充的字母数 = 原序列S的长度 —  S和e的最长公共子串长度

其中e为s的逆序字符串。

于是这题就变成了求s和e的最长公共子序列的长度。

dp[ i ][ j ]表示 s 序列的前 i 位与 e 序列的前 j 位的最长公共子序列的长度

动规方程为如果s[ i-1 ]==e[ j-1 ]，那么dp[ i ][ j ]=dp[ i-1 ][ j-1 ]+1

如果不相等dp[ i ][ j ]=max(dp[ i-1 ][ j ],dp[ i ][ j-1])

这道题的字符串范围是5000，那么定义二维数组是会空间超限的，所以不能定义dp[ 5000 ][ 5000 ]

再看动态方程，会发现，对于 i 这个变量我们只用到了 i 和 i-1 。那么我们可以把它们用0和1表示，是不会有重复的；所以我定义的dp[ 2 ][ 5000] ;对每一个 i 对2取余( i % 2 )；

同样对于 j 这个变量也可以用同样的方法。

AC代码

#include<iostream>#include<cstdio>#include<cstring>using namespace std;char s[5005],e[5004];//s存入字符串，e为s的倒序 int dp[2][5005];int main(){int n;while(scanf("%d",&n)!=EOF){scanf("%s",s);for(int i=0;i<n;i++){e[i]=s[n-i-1];}memset(dp,0,sizeof(dp));int ans=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(s[i-1]==e[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]);}ans=max(dp[i%2][j],ans);}}printf("%d\n",n-ans);}return 0;}