POJ - 1458 Common Subsequence —— 最长公共子序列

最新推荐文章于 2020-07-04 14:56:09 发布

Lngxling

最新推荐文章于 2020-07-04 14:56:09 发布

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本文介绍了一种使用动态规划解决最长公共子序列问题的方法，通过两个字符串作为输入，输出两者之间的最长公共子序列的长度。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 55054		Accepted: 22947

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

题意：找最长上升子序列只需按顺序即可

思路：dp[i][j]表示前一个串的前i个字母和后一个串的前j个字母的最长公共子序列的长度

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#define max_ 1010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int dp[max_][max_];

int main(int argc, char const *argv[])
{
	char s1[max_],s2[max_];
	while(scanf("%s%s",s1,s2)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		int i,j;
		for(i=1;i<=strlen(s1);i++)
		{
			for(j=1;j<=strlen(s2);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]);
			}
		}
		printf("%d\n",dp[strlen(s1)][strlen(s2)]);
	}
	return 0;
}