POJ 1458（最长公共子序列）

最新推荐文章于 2019-06-13 07:58:44 发布

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#POJ 1458 #最长公共子序列

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本文介绍了解决最长公共子序列(LCS)问题的方法，包括定义、输入输出格式及样例，并通过C++代码实现了动态规划算法来寻找两字符串的最长公共子序列长度。

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48407		Accepted: 19946

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

分析：

第一步就是找子问题，也就是把一个大的问题分解成子问题。这里我们设两个字符串A、B，A = "a0, a1, a2, ..., am-1"，B = "b0, b1, b2, ..., bn-1"。

（1）如果am-1 == bn-1，则当前最长公共子序列为"a0, a1, ..., am-2"与"b0, b1, ..., bn-2"的最长公共子序列与am-1的和。长度为"a0, a1, ..., am-2"与"b0, b1, ..., bn-2"的最长公共子序列的长度+1。
（2）如果am-1 != bn-1，则最长公共子序列为max（"a0, a1, ..., am-2"与"b0, b1, ..., bn-1"的公共子序列，"a0, a1, ..., am-1"与"b0, b1, ..., bn-2"的公共子序列）
引入一个二维数组c[][]，其中c[i][j]记录X[i]与Y[j]的LCS长度
这样我们可以总结出该问题的递归形式表达：

代码：

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

typedef long long int ll;

const int maxn = 1005;
char s1[maxn];
char s2[maxn];
int a[maxn][maxn];

int main(){
	int n,m;
	while(~scanf("\n%s\n%s",s1+1,s2+1)){
		n = strlen(s1+1);
		m = strlen(s2+1);
		memset(a,0,sizeof(a));
		for(int i=1; i<=n; i++)
			for(int j=1; j<=m; j++){
				if(s1[i]==s2[j])
					a[i][j] = a[i-1][j-1]+1;
				else
					a[i][j] = max(a[i-1][j],a[i][j-1]);
			}
		printf("%d\n",a[n][m]);
	}
	
	return 0;
}