zoukankan      html  css  js  c++  java
  • POJ 1159, Palindrome

    Time Limit: 3000MS  Memory Limit: 65536K
    Total Submissions: 26714  Accepted: 8915


    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
    5
    Ab3bd

    Sample Output
    2

    Source
    IOI 2000


    // POJ1159.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <algorithm>
    using namespace std;

    int main(int argc, char* argv[])
    {
        
    int N;
        scanf(
    "%d\n"&N);

        
    char word[5001];
        gets(word);

        
    int DP[2][5001];
        memset(DP,
    0,sizeof(DP));

        
    for (int i = 1; i <= N; ++i)
        {
            DP[i 
    & 1][0= 0;
            
    for (int j = 1; j <= N; ++j)
            {
                
    if (word[i-1]==word[N-j])DP[i & 1][j] = DP[(i-1& 1][j-1+ 1;
                
    else DP[i & 1][j] = max(DP[(i-1& 1][j], DP[i & 1][j-1]);
            }
        }
        
        cout 
    << N - DP[N&1][N] << endl;

        
    return 0;
    }

  • 相关阅读:
    HDOJ 1846 Brave Game
    并查集模板
    HDU 2102 A计划
    POJ 1426 Find The Multiple
    POJ 3278 Catch That Cow
    POJ 1321 棋盘问题
    CF 999 C.Alphabetic Removals
    CF 999 B. Reversing Encryption
    string的基础用法
    51nod 1267 4个数和为0
  • 原文地址:https://www.cnblogs.com/asuran/p/1583117.html
Copyright © 2011-2022 走看看