动态规划—杭电1159 Common Subsequence

最新推荐文章于 2021-07-15 17:16:01 发布

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最新推荐文章于 2021-07-15 17:16:01 发布

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本文介绍了一个经典的计算机科学问题——寻找两个字符串的最长公共子序列，并提供了一种使用动态规划解决该问题的方法。通过示例展示了算法的运行过程及实现代码。

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http://acm.hdu.edu.cn/showproblem.php?pid=1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9101 Accepted Submission(s): 3681

Problem 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
abcfbc abfcab
programming contest
abcd mnp

Sample Output
4
2
0

辅助空间变化示意图

	a	b	c	f	b	c
a	1	1	1	1	1	1
b	1	2	2	2	2	2
f	1	2	2	3	3	3
c	1	2	3	3	3	4
a	1	2	3	3	3	4
b	1	2	3	3	4	4

#include <iostream>
#include <string.h>
#define max(a,b) a>b?a:b
using namespace std;
char a[1001],b[1001];
int dp[1001][1001];
int main()
{
	while(cin>>a>>b)
	{
		memset(dp,0,sizeof(dp));
		int lena,lenb;
		lena=strlen(a);
		lenb=strlen(b);
		int i,j;
		for(i=0;i<lena;i++)
			for(j=0;j<lenb;j++)
			{
				if(a[i]==b[j])
						dp[i+1][j+1]=dp[i][j]+1;
				else
						dp[i+1][j+1] = max(dp[i+1][j],dp[i][j+1]);
			}
			cout<<dp[lena][lenb]<<endl;				
	}
	return 0;
}