POJ 1458 Common Subsequence（最长公共子序列问题）

Common Subsequence

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_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.

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

本题关键在于状态转移方程，DP水题，直接贴代码

递归代码如下：

#include<iostream>
#include<string>
using namespace std;
int dp[1000][1000];//记忆化数组
string s1,s2;
int max(int a,int b)
{return a>b?a:b;}
int maxlen(int i,int j)
{
	if(dp[i][j]!=-1)
		return dp[i][j];
	if(!(i&&j))
		return 0;
	if(s1[i-1]==s2[j-1])
		dp[i][j]=maxlen(i-1,j-1)+1;
	else
		dp[i][j]=max(maxlen(i-1,j),maxlen(i,j-1));
	return dp[i][j];
}
int main()
{

	while(cin>>s1>>s2)
	{
		int len1=s1.length(),len2=s2.length();
		memset(dp,-1,sizeof(dp));
		maxlen(len1,len2);
		cout<<dp[len1][len2]<<endl;
	}
	return 0;
}

递推代码：

#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;
int i,j;
for( i = 0;i <= length1; i ++ )
maxLen[i][0] = 0;
for( j = 0;j <= length2; j ++ )
maxLen[0][j] = 0;
55
for( i = 1;i <= length1;i ++ ) {
for( 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;
}

//空间优化一维：（WA，哪位道友帮找下哪里错了这个。。）

#include<iostream>
#include<string>
using namespace std;
int dp[1000];
string s1,s2;
int max(int a,int b)
{return a>b?a:b;}
int maxlen()
{
	int i,j,ans=0;
	int len1=s1.length(),len2=s2.length();
			memset(dp,0,sizeof(dp));
	for(i=0;i<len1;i++)
	{
		for(j=0;j<len2;j++)
			if(s1[i]==s2[j])
				dp[j+1]=dp[j]+1;
			else
				dp[j+1]=max(dp[j],dp[j+1]);
			ans=ans>dp[len2]?ans:dp[len2];
	}
	return ans;
}
int main()
{
	while(cin>>s1>>s2)
	{
		cout<<maxlen()<<endl;
	}
	return 0;
}