LCS简单问题

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本文介绍了一个经典的计算机科学问题——寻找两个字符串之间的最长公共子序列的长度，并提供了一段C++代码实现。该问题通常应用于生物信息学、文本比较等领域。

问题 B: 选拔赛专用题(2)

时间限制: 1 Sec 内存限制: 128 MB
提交: 52 解决: 14
[ 提交][ 状态][ 讨论版]

题目描述

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 ij = 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.

输入

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.

输出

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.

样例输入

abcfbc         abfcab
programming    contest 
abcd           mnp

样例输出

4
2
0

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

char s1[1000],s2[1000];
int dp[1000][1000];
int len1,len2;

void LCS()
{
int i,j;
memset(dp,0,sizeof(dp));
for(i = 1; i<=len1; i++)
{
for(j = 1; j<=len2; j++)
{
if(s1[i-1] == s2[j-1])
dp[i][j] = dp[i-1][j-1]+1;
else
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
}
int main()
{
while(cin>>s1>>s2)
{
len1 = strlen(s1);
len2 = strlen(s2);
LCS();
cout<<dp[len1][len2]<<endl;
}
return 0;
}