zoukankan      html  css  js  c++  java
  • Common Subsequence

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 53247   Accepted: 22084

    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

    Source

     
    用dp[n][m]表示第一个喘选到n,第二个串选到m
     
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<algorithm>
     6 using namespace std;
     7 const int MAXN=601;
     8 const int maxn=0x7fffff;
     9 char a[MAXN];
    10 char b[MAXN];
    11 int dp[MAXN][MAXN];
    12 int la,lb;
    13 int main()
    14 {
    15     while(~scanf("%s%s",a,b))
    16     {
    17         la=strlen(a);
    18         lb=strlen(b);
    19         memset(dp,0,sizeof(dp));
    20         //dp[0][0]=1;
    21         for(int i=1;i<=la;i++)
    22         {
    23             for(int j=1;j<=lb;j++)
    24             {
    25                 
    26                 if(a[i-1]==b[j-1])
    27                     dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
    28                 else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    29             }
    30         }
    31         printf("%d
    ",dp[la][lb]);
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    《观止》读后感
    读产品经理相关书籍有感
    windows phone 7基础点随手记
    Windows phone 7画图画字
    《CLR via C#》读书笔记
    《Beginning C# Objcets》学习笔记
    ASP.NET 使用alert弹出对话框后,CSS样式失效,字体变大的解决方法
    .NET COOKIE /SESSION/CACHE操作类
    【原创】VS2005 Web应用程序打包并安装数据库
    存储过程实现多条件查询
  • 原文地址:https://www.cnblogs.com/zwfymqz/p/7202106.html
Copyright © 2011-2022 走看看