Common Subsequence poj-1458 动态规划

最新推荐文章于 2020-07-04 14:56:09 发布

原创最新推荐文章于 2020-07-04 14:56:09 发布 · 270 阅读

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该博客主要介绍了如何使用动态规划解决POJ-1458问题，即找到两个字符串的最长公共子序列。通过读取输入的两个字符串，计算它们的长度，并用二维数组maxlen记录每个位置上最长公共子串的长度。最后输出最长公共子序列的长度。代码中包含了详细的解释和边界条件处理。

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 57168		Accepted: 23882

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

#include<iostream>
#include<string>
#include<string.h>
using namespace std;

char sz1[1000];//存放第一个字符串
char sz2[1000];//存放第二个字符串
int maxlen[1000][1000];//存放公共子串的长度
//maxlen[i][j]表示s1的左边i个长度和s2
的左边j个长度的子串组成的最长公共子串的长度
int main()
{
while(cin>>sz1)//出入数据
{
cin>>sz2;
int len1=strlen(sz1);//测量长度
int len2=strlen(sz2);
int i,j;
for(i=0; i<len1; i++)
{
maxlen[i][0]=0;//边界条件
}
for(j=0; j<len2; j++)
{
maxlen[0][j]=0;
}
for(i=1; i<=len1; i++)
{
for(j=1; j<=len2; j++)
{
if(sz1[i-1]==sz2[j-1])//如果相等的话，从s[0]开始算
{
maxlen[i][j]=maxlen[i-1][j-1]+1;
}
else
{//自己实验下找到规律
maxlen[i][j]=max(maxlen[i-1][j],maxlen[i][j-1]);
}
}
}
cout<<maxlen[len1][len2]<<endl;
}
return 0;
}