zoukankan      html  css  js  c++  java
  • HDU--1159 Common Subsequence

      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, x ij = 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:求最长相同子序列
    代码:
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 using namespace std;
     5 int dp[1005][1005]={0};
     6 int main()
     7 {
     8     char s1[10000];
     9     char s2[10000];
    10     s1[0]='0';
    11     s2[0]='0';
    12     
    13     while(scanf("%s %s",s1+1,s2+1)!=EOF)
    14     {
    15         
    16         int length1=strlen(s1)-1;
    17         int length2=strlen(s2)-1;
    18         for(int i=1;i<=length1;i++)
    19         {
    20             for(int j=1;j<=length2;j++)
    21             {
    22                 if(s1[i]==s2[j])
    23                 {
    24                     dp[i][j]=dp[i-1][j-1]+1;
    25                 }
    26                 else
    27                 {
    28                     dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    29                 }
    30             }
    31         }
    32         printf("%d
    ",dp[length1][length2]);
    33         
    34     }
    35     return 0;
    36 }

    代码:

    代码:
  • 相关阅读:
    sourcenav安装
    vim-addon-manager【转】
    zmq重点
    一个简单的解决方法:word文档打不开,错误提示mso.dll模块错误。
    微软验证码项目 Captcha Code Demo 从 .NET Core 1.1.2升级到2.1.0
    海康设备如何接入萤石开放平台
    ABP 启用多租户实现数据隔离
    Docker 开发者常用操作命令
    .NET Core 深度克隆对象,引用类型的平行世界
    详解 .NET Core 遍历 List 并移除项
  • 原文地址:https://www.cnblogs.com/hss-521/p/7305409.html
Copyright © 2011-2022 走看看