[dp]poj1458 -Common Subsequence(LCS)

最新推荐文章于 2021-09-12 22:21:57 发布

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最新推荐文章于 2021-09-12 22:21:57 发布

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分类专栏： ————ACM训练———— 2、动态规划

本文链接：https://blog.youkuaiyun.com/tlonline/article/details/46128069

————ACM训练———— 同时被 2 个专栏收录

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2、动态规划

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本文介绍了一个经典的计算机科学问题——最长公共子序列问题，并提供了一种基于动态规划的解决方案。通过实例输入输出展示了如何求解两个字符串的最长公共子序列长度。

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41846		Accepted: 16868

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, x _{i_j} = 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

一个裸的最长公共子序列问题。。没什么好说的。。用最简单的dp做就能过了。点击这里查看实现LCS的方式。。

#include<iostream>
#include<cstdio>
#include<set>
#include<cstring>
using namespace std;

const int MAXN =1001;
string s,t;
int dp[MAXN][MAXN];

int main()
{
    while(cin>>s>>t)
    {
        int n = s.length();
        int m = t.length();
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(s[i] == t[j])
                    dp[i+1][j+1] = dp[i][j] + 1;
                else
                    dp[i+1][j+1] = max(dp[i][j+1],dp[i+1][j]);
            }
        }
            cout<<dp[n][m]<<endl;
    }
}