POJ 1458 Common Subsequence【最长公共子序列】_已知x=abcfbc,y=abfcab.求x和y的最长公共子序列。-优快云博客

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，并提供了一种有效的解决方案。通过使用动态规划的方法，文章详细解释了如何找出两个给定序列的最长公共子序列的长度。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 59599		Accepted: 24845

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

题意：给你两个序列，让你求他们的最长公共子序列的长度。

值得注意的是：公共子序列不是公共子串。

列如：X={3,0,2,5,1,7,6,4}，它的子序列可以为{3,5,4}。但其子串为位置相临近的几个元素，如{2,5,1,7}。

最长公共子序列的求法参看网址：

https://blog.youkuaiyun.com/hrn1216/article/details/51534607

思路：假设我们用c[i][j]表示Xi和Yi的最长公共子序列的长度，则有如下递推公式：

题解：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
#define M(a,b) memset(a,b,sizeof(a))
const int MAXN=1005;
int c[MAXN][MAXN];
int main()
{
    char s1[MAXN];
    char s2[MAXN];
    while(~scanf("%s",&s1))
    {
        scanf("%s",&s2);
        int n=strlen(s1);
        int m=strlen(s2);
        M(c,0);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {

                    if(s1[i]==s2[j])///字符s1[i]与s2[j]相等，c[i][j]= c[i-1,j-1] + 1。
                           c[i+1][j+1]=c[i][j]+1;
                 else
                    c[i+1][j+1]=max(c[i][j+1],c[i+1][j]);///否则，取c[i-1,j] 和 c[i,j-1]的最大值
            }

        }
        printf("%d\n",c[n][m]);
    }
    return 0;
}