Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57902 Accepted Submission(s): 26737

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

理解题意：

首先我们的了解什么是子序列，一个给定序列的子序列是在该序列中删去若干元素后的得到的序列，可以是不连续的。

做题思路：

找到两个序列的最长公共子序列，用最基础的DP模板即可

代码：

#include <bits/stdc++.h>

using namespace std;

int dp[][];

string str1,str2;

int LCS(){

    for(int i = ;i < str1.length(); i++){

        for(int j = ;j < str2.length(); j++){

            if(str1[i] == str2[j])

                dp[i+][j+] = dp[i][j] + ;

            else

                dp[i+][j+] = max(dp[i][j+],dp[i+][j]);

        }

    }

    return dp[str1.length()][str2.length()];

} 

int main()

{

    while(cin>>str1>>str2){

        cout<<LCS()<<endl;

    }

    return ;

}

巴特西

最长公共子序列-Hdu1159

Common Subsequence

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