HDOJ1513 LCS水题

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6434    Accepted Submission(s): 2128

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

 

Source

IOI 2000

 

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不难看出规律，构造一个和原串序列相反的字符串，然后求LCS，最后用总长度n-lcs就是答案了。

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;const int maxn = 5005;char a[maxn],b[maxn];int i,j,k,n,dp[2][maxn];bool flag;void init(){    scanf("%s",a);    for (i=0; i<n; i++) b[i] = a[n-1-i];    memset(dp,0,sizeof(dp));}int main(){    while (scanf("%d",&n)!=EOF){        init();        for (i=1; i<=n; i++) {            for (j=1; j<=n; j++) {                if (a[i-1]==b[j-1]) dp[1][j] = dp[0][j-1] + 1;                else dp[1][j] = max(dp[0][j],dp[1][j-1]);            }            for (k=0; k<=n; k++) dp[0][k] = dp[1][k];        }        printf("%d\n",n-dp[1][n]);    }    return 0;}