Poj 1458 Common Subsequence【LCS】【dp】

mengxiang000000

于 2016-08-01 09:31:20 发布

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分类专栏： dp 文章标签： Poj 1458 PKU 1458

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本文介绍了一种解决最长公共子序列问题的有效算法。通过动态规划方法，文章详细阐述了如何计算两个字符串之间的最长公共子序列长度，并给出了完整的C++实现代码。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47612		Accepted: 19568

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

Source

Southeastern Europe 2003

题目大意：

给你两个字符串，求最长公共子序列的长度.

思路：

1、设定dp【i】【j】表示第一个字符串以第i位结尾，第二个字符串以第j位结尾的最长公共子序列长度。

2、辣么状态转移方程：

①if(a【i】==b【j】)dp【i】【j】=dp【i-1】【j-1】+1；

②else dp【i】【j】=max（dp【i-1】【j】，dp【i】【j-1】）；

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char a[100000];
char b[100000];
int dp[1005][1005];
int main()
{
    while(~scanf("%s%s",a,b))
    {
        int n1,n2;
        n1=strlen(a);
        n2=strlen(b);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n1;i++)
        {
            for(int j=1;j<=n2;j++)
            {
                if(a[i-1]==b[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i][j-1],dp[i][j]);
                    dp[i][j]=max(dp[i-1][j],dp[i][j]);
                }
            }
        }
        printf("%d\n",dp[n1][n2]);
    }
}