zoukankan      html  css  js  c++  java
  • POJ-1458 Common Subsequence (最长公共子序列)

    Common Subsequence
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 39851   Accepted: 16030

    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
    大致题意:求两个字符串的最长公共子序列的长度。

    思路:这是一个很经典的最长公共子序列(LCS)问题。动态转移方程式如下,设有字符串X和字符串Y,dp[i,j]表示的是X的钱i个字符与Y的钱j个字符的最长公共子序列的长度。如果X[i]==Y[j],那么这个字符与之前的LCS一定可以构成一个新的LCS;如果X[i]!=Y[j],则分别考查dp[i-1,j]和dp[i][j-1],选择其中的较大者为LCS。


     1 #include<iostream>
     2 #include<cstdio>
     3 #include<algorithm>
     4 #include<cmath>
     5 #include<string>
     6 using namespace std;
     7 const int MAX = 500;
     8 int dp[MAX][MAX] ={0};
     9 int main()
    10 {
    11     int len1,len2;
    12     string str1,str2;
    13     while(cin>>str1>>str2)
    14     {
    15         len1=str1.length();
    16         len2=str2.length();
    17         for(int i=1;i<=len1;i++)
    18         {
    19             for(int j=1;j<=len2;j++)
    20             {
    21                 if(str1[i-1]==str2[j-1])
    22                 {
    23                     dp[i][j]=dp[i-1][j-1]+1;
    24                 }
    25                 else
    26                 {
    27                     dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    28                 }
    29             }
    30         }
    31         cout<<dp[len1][len2]<<endl;
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    记intouch SMC local下驱动丢失问题解决
    随机变量的频数分布图
    Clonal hematopoiesis of indeterminate potential(意义不明的克隆性造血)-CHIP
    聚类热图
    linux把软件安装到指定目录
    Phred-scale quality scores的相关内容
    GATK的硬过滤
    VCF和GVCF格式说明
    非root权限 安装更新gcc
    在非小细胞肺癌中,MET基因的14号外显子突变和年龄,依赖于癌症阶段的CNV,C-MET过表达的关系
  • 原文地址:https://www.cnblogs.com/caterpillarofharvard/p/4225815.html
Copyright © 2011-2022 走看看