HDOJ 1159 Common Subsequence【模板题】

最新推荐文章于 2020-04-13 10:58:43 发布

原创最新推荐文章于 2020-04-13 10:58:43 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

LCS 专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个经典的计算机科学问题——最长公共子序列问题，并提供了一段完整的C++代码实现。该问题要求找出两个字符串之间的最长公共子序列的长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34223 Accepted Submission(s): 15602

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

Source

Southeastern Europe 2003

Recommend

Ignatius

LCS模板题，下面贴上模板。

只要理解LCS，然后这题就是个模板题，没有难度。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char ch1[1111],ch2[1111];
int dp[1111][1111];
int main()
{
	int i,j;
	while(~scanf("%s%s",ch1,ch2))
	{
		memset(dp,0,sizeof(dp));
		int l1=strlen(ch1);
		int l2=strlen(ch2);	
		for(i=1;i<=l1;i++)
		{
			for(j=1;j<=l2;j++)
			{
				if(ch1[i-1]==ch2[j-1])
					dp[i][j]=dp[i-1][j-1]+1;//相等的话+1
				else
					dp[i][j]=max(dp[i-1][j],dp[i][j-1]);//不相等选之前状态下最长的那个
			}
		}
		printf("%d\n",dp[l1][l2]);
	}
	return 0;
}