POJ 1458（最长公共子序列）

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#POJ 1458 #最长公共子序列

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本文讲解了如何使用动态规划解决最长公共子序列问题，并提供了一段实现代码。文章首先定义了子序列的概念，随后详细解释了状态转移方程，最后通过示例说明了算法的应用。

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Common Subsequence

Time Limit: 1000MS

Memory Limit: 10000K

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

问题分析

题意：求俩个字符串的最长公共子串的长度
一道经典dp。
在解决这一类的问题时，我们所需要做的肯定是
⒈找子问题
⒉确定状态
⒊找出状态转移方程
那怎么找子问题呢？题意要求两个字符串的最长公共子串，不如试着去求s1左边0->s1.length-1和s2左边0->s2.length-1的的最长公共部分。也就是在0->s1.length-1和0->s2.length-1中求maxLength(i,j)，因此，maxLength(i,j)就是本题的状态。所以我们可以通过去比较s1[i-1]和s2[j-1],来递推maxLength(i,j)。所以本题的状态方程为
if(s1[i-1]==s2[j-1])
maxLength(i,j) = maxLength(i-1,j-1)
else
maxLength(i,j) = max(maxLength(i-1,j),maxLength(i,j-1))
【s1[i-1]!= s2[j-1]时，maxLenth(s1,s2)不会比maxLength(s1,s2j-1) 和MaxLen(s1i-1,s2)两者之中任何一个小，也不会比两者都大。】
最后输出maxLength(s1.length,s2.length)即可。

好，下面上ACcode（^_^）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
const int inf = 0x3f3f3f;
const int N = 500;
using namespace std;

int dp[N][N];

int main()
{
    char a[N],b[N];
    while(~scanf(" %s %s",a ,b))
    {
        int a1 = strlen(a);
        int b1 = strlen(b);
        for(int i = 1; i <= a1; i++)
        {
            for(int j = 1; j <= b1; ++j)
            {
                if(a[i-1]==b[j-1])
                dp[i][j] = dp[i-1][j-1]+1;
                else
                dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
            }
        }
        cout<<dp[a1][b1]<<endl;
    }
    return 0;
}