poj 1458 Common Subsequence

最长公共子序列问题求解

weixin_30247159

于 2011-11-21 18:51:00 发布

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/21/2257538.html

本文介绍了一个经典的计算机科学问题——寻找两个字符串序列的最长公共子序列，并提供了一段C++实现代码。该问题通过动态规划算法解决，旨在帮助读者理解如何高效地找出两组数据之间的相似性。

Common Subsequence

Time Limit: 1000MS
Memory Limit:10000K

Total Submissions: 26378
Accepted:10221

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

#include <iostream>

#include <string>

using namespace std;

string str1;

string str2;

int num[256][256];

int lcs ( string s1 , string s2 , int m , int n )

	int i, j;

	for ( i=1 ; i <= m ; i++ )

		for ( j = 1 ; j <= n ; j++ )

			if ( s1[i-1] == s2[j-1] )

				num[i][j] = num[i-1][j-1] + 1;

			else

				if ( num[i][j-1] < num[i-1][j] )

					num[i][j] = num[i-1][j];

				else

					num[i][j] = num[i][j-1];

	return num[m][n];

int main()

	int len1, len2;

	while ( cin>>str1>>str2 )

		memset ( num , 0 , sizeof(num) );

		len1 = str1.length();

		len2 = str2.length();

		int out = lcs(str1,str2,len1,len2);

		cout<<out<<endl;

	return 0;

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2257538.html