zoukankan      html  css  js  c++  java
  • 1159 Palindrome

    Palindrome
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 68562   Accepted: 23869

    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

    用的LIS的做法 参考1458

    字符串s长度为N 将输入的字符串倒过来记做rs 则N - (s与rs的最长公共子序列的长度) 就是答案

    用scanf或者getchar()都是1600MS 不知道0MS的是怎么做的

    #include <stdio.h>
    #include <iostream>
    #include <cstring>
    #include <vector>
    #include <algorithm>
    const int si = 5016;
    using namespace std;
    char s[si], rs[si];
    int dp[2][si];
    int main() {
        int N;
        cin >> N;
        scanf("%s", s);
        for (int i = 0; i < N; i++) rs[N - i - 1] = s[i];
        int e = 0;
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= N; j++) {
                dp[e][j] = max(dp[1 - e][j], dp[e][j - 1]);
                if (s[i - 1] == rs[j - 1]) {
                    dp[e][j] = max(dp[e][j], dp[1 - e][j - 1] + 1);
                }
            }
            e = 1 - e;
        }
        cout << N - dp[1 - e][N];
        return 0;
    }
  • 相关阅读:
    大话设计模式Python实现-代理模式
    大话设计模式Python实现-装饰模式
    大话设计模式Python实现-策略模式
    设计模式Python实现-简单工厂模式
    Python文件读写机制
    python 多线程剖析
    I/O多路复用-EPOLL探索
    Python学习笔记:魔术方法详解
    Django学习笔记:为Model添加Action
    【Django】Django Debug Toolbar调试工具配置
  • 原文地址:https://www.cnblogs.com/smatrchen/p/10585761.html
Copyright © 2011-2022 走看看