poj1159

Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 46820 Accepted: 16005

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

比较水的dp，具体是比较原串和反串的最大公共子序列。

貌似不知道为啥G++就过不去。。。

另外需要注意下数组空间...滚动数组 - -||

#include <iostream>#include <algorithm>#include <cstdio>#define MAX 5001using namespace std;int main(){    int len;    cin>>len;    char s1[MAX],s2[MAX];    scanf("%s",s1);    for(int i=len-1;i>=0;i--)        s2[len-1-i] = s1[i];    short cost[2][MAX];    short comp = -1;    for(int i=1;i<=len;i++)        for(int j=1;j<=len;j++)        {            if(s1[i-1]==s2[j-1]){                cost[i%2][j] = cost[(i-1)%2][j-1]+1;                comp = max(comp,cost[i%2][j]);            }            else                cost[i%2][j] = max(cost[(i-1)%2][j],cost[i%2][j-1]);        }    cout<<len-comp<<endl;    return 0;}