zoukankan      html  css  js  c++  java
  • poj1458——dp,lcs

    poj1458——dp,lcs

    Common Subsequence
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 40529   Accepted: 16351

    Description

    A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

    Input

    The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

    Output

    For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

    Sample Input

    abcfbc         abfcab
    programming    contest 
    abcd           mnp

    Sample Output

    4
    2
    0
    题意:求lcs,水题!
    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    
    using namespace std;
    
    const int maxn=1000100;
    char s[maxn],t[maxn];
    int dp[2][maxn];
    
    int main()
    {
        while(scanf("%s%s",s,t)!=EOF){
            int ls=strlen(s),lt=strlen(t);
            memset(dp,0,sizeof(dp));
            char *ss=s-1,*tt=t-1;
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=ls;i++){
                for(int j=1;j<=lt;j++){
                    if(ss[i]==tt[j]) dp[i%2][j]=dp[(i+1)%2][j-1]+1;
                    else dp[i%2][j]=max(dp[(i+1)%2][j],dp[i%2][j-1]);
                }
            }
            printf("%d
    ",dp[ls%2][lt]);
        }
        return 0;
    }
    View Code
    没有AC不了的题,只有不努力的ACMER!
  • 相关阅读:
    Python PyQt5 Pycharm 环境搭建及配置
    Python Appium 元素定位方法简单介绍
    unittest单元测试简单介绍
    什么是multipart/form-data请求
    new HttpClient().PostAsync封装参数
    httpPostedFile实现WEBAPI文件上传
    Asp.Net WebApi接口返回值IHttpActionResult
    Asp.Net WebApi上传图片
    如果项目在IIS发布后,出现System.ComponentModel.Win32Exception: 拒绝访问。
    C# ASP.NET 控制windows服务的 开启和关闭 以及重启
  • 原文地址:https://www.cnblogs.com/--560/p/4356620.html
Copyright © 2011-2022 走看看