HDU/HDOJ 1159/POJ 1458 Common Subsequence（最长公共子序列LCS，滚动数组）

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本文介绍了一个经典的计算机科学问题——最长公共子序列(LCS)，并提供了两种不同的动态规划解决方案，包括传统的二维数组方法和优化后的滚动数组方法。

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

dd poj>>>>> http://poj.org/problem?id=1458

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47893		Accepted: 19715

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

Source

Southeastern Europe 2003

题意：

给定两个字符串，求出两者最长的公共上升子序列（LCS）。

思路：

裸露的模板题。

Code1（滚动数组的优化）：

/*
	Name:
	Copyright:
	Author:Shyazhut
	Date: 12-08-16 15:30
	Description: LCS的DP优化，滚动数组的引入
*/
#include<cstdio>
#include<cstring>
const int MYDD=1103;

int dp[2][MYDD];
char a[MYDD],b[MYDD];
int MAX(int x,int y){
	return x>y? x:y;
}

int main() {
	while(scanf("%s%s",a,b)!=EOF) {
		memset(dp,0,sizeof(dp));
		int lena=strlen(a);
		int lenb=strlen(b);
		for(int j=1; j<=lena; j++) {
			for(int k=1; k<=lenb; k++) {
				if(a[j-1]==b[k-1]) {
					dp[j%2][k]=dp[(j-1)%2][k-1]+1;
				} else {
					dp[j%2][k]=MAX(dp[(j-1)%2][k],dp[j%2][k-1]);
				}
			}
		}
		printf("%d\n",dp[lena%2][lenb]);
	}
	return 0;
}
/*By: Shyazhut*/

Code2（无滚动数组的优化）：

/*
	Name:poj Common Subsequence
	Copyright:
	Author:Shyazhut
	Date: 12-08-16 15:15
	Description: LCS的学习
*/
#include<cstdio>
#include<cstring>
const int MYDD=1103;

char a[MYDD],b[MYDD];
int dp[MYDD][MYDD];
int MAX(int x,int y){
	return x>y? x:y;
}

int main() {
	while(scanf("%s%s",a,b)!=EOF) {
		memset(dp,0,sizeof(dp));
		int lena=strlen(a);
		int lenb=strlen(b);
		for(int j=1; j<=lena; j++) {
			for(int i=1; i<=lenb; i++) {
				if(a[j-1]==b[i-1]) {
					dp[j][i]=dp[j-1][i-1]+1;
				} else {
					dp[j][i]=MAX(dp[j-1][i],dp[j][i-1]);
				}
			}
		}
		printf("%d\n",dp[lena][lenb]);
	}
	return 0;
}
/*By: Shyazhut*/