动态规划之Common Subsequence（ACM）

最新推荐文章于 2020-09-22 23:59:10 发布

原创最新推荐文章于 2020-09-22 23:59:10 发布 · 744 阅读

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#ACM #动态规划

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本文介绍了一种解决最长公共子序列问题的算法，通过动态规划的方法，输入两字符串序列，输出它们的最长公共子序列长度。代码实现简洁，直接给出AC代码。

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 Copy

abcfbc abfcab
programming contest 
abcd mnp

Sample Output Copy

4
2
0

废话不多说，直接上AC代码：

#include<iostream>
#include<cstring>
using namespace std;
char sz1[1000];
char sz2[1000];
int maxLen[1000][1000];
int main()
{
	while(cin>>sz1>>sz2)
	{
		int length1=strlen(sz1);
		int length2=strlen(sz2);
		int nTmp;
		for(int i=0;i<=length1;i++)
		maxLen[i][0]=0;
		for(int j=0;j<=length2;j++)
		maxLen[0][j]=0;
		for(int i=1;i<=length1;i++)
		{
			for(int j=1;j<=length2;j++)
			{
			if(sz1[i-1]==sz2[j-1])
			maxLen[i][j]=maxLen[i-1][j-1]+1;
			else
			maxLen[i][j]=
			max(maxLen[i][j-1],maxLen[i-1][j]);
			}
		}
		cout<<maxLen[length1][length2]<<endl;
	}
	return 0; 
 }