zoukankan      html  css  js  c++  java
  • poj1159 Palindrome

    Palindrome
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 44186   Accepted: 15050

    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

    输入a串,求其变成回文串所要进行的插入或删除操作的最少的步骤

    Dp[i][j] 为a[i~j] 子串变为回文串所要进行插入或删除的操作的最少步骤。

                Min{dp[i][j-1]+1, dp[i-1][j]+1}       (a[i]!=a[j])

    Dp[i][j]=

                Min{dp[i][j-1]+1,dp[i-1][j]+1,dp[i-1][j-1]}   (a[i]==a[j])

    则最后答案为dp[1][n]

    
    
    #include<cstdio>
    #include<string>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    short dp[5001][5001];
    int n;
    string s;
    int main(){
        while(cin>>n>>s){
            memset(dp,0,sizeof(dp));
            for(int i=n-1;i>=0;i--){
                 for(int j=i;j<n;j++){
                    if(s[i]==s[j])dp[i][j]=dp[i+1][j-1];
                    else
                    dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;
                 }
            }
                 printf("%d
    ",dp[0][n-1]);
        }
      return 0;
    }
    View Code
  • 相关阅读:
    简单错误记录
    识别有效的IP地址和掩码并进行分类统
    爬虫必备—BeautifulSoup
    爬虫必备—requests
    Shellinabox安装及使用教程
    Django——REST framework
    SaltStack部署
    使用js在HTML中自定义字符串格式化方法
    3种上传图片并实现预览的方法
    Ajax
  • 原文地址:https://www.cnblogs.com/demodemo/p/4732574.html
Copyright © 2011-2022 走看看