PKU-1458 Common Subsequence (最长公共子序列LCS)

最新推荐文章于 2016-08-01 09:31:20 发布

gk1988

最新推荐文章于 2016-08-01 09:31:20 发布

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分类专栏： ACM-DP

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原题链接 http://poj.org/problem?id=1458

Common Subsequence

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 Code

/* 
  最长公共子序列 
  DP[i, j] = 0  if i == 0 or j == 0; 
  DP[i, j] = DP[i - 1][j - 1] + 1;  if ch1[i] == ch2[j] 
  DP[i, j] = max(DP[i - 1][j], DP[i][j - 1]);  if ch1[i] != ch2[j] 
 
  滚动数组： 
  dp[i][j]的值只依赖于dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]; 
  dp[i%2][j]=max(dp[(i-1)%2][j], dp[i%2][j-1]); 
  注意上面的取余运算，重复利用这2行数组，数组好象在“滚动”一样，所以叫滚动数组 
*/  


#include <iostream>
using namespace std;

#define Max 1001

int match[2][Max];
char input1[Max], input2[Max];

int main() {
	int i, j;
	while (scanf("%s %s", input1 + 1, input2 + 1) != EOF) {
		
		memset(match, 0, sizeof(match));

		for (i = 1; input1[i] != '\0'; i++) {
			for (j = 1; input2[j] != '\0'; j++) {
				if (input1[i] == input2[j]) {
					match[i % 2][j] = match[(i - 1) % 2][j - 1] + 1;
				} else {
					if (match[(i - 1) % 2][j] > match[i % 2][j - 1]) {
						match[i % 2][j] = match[(i - 1) % 2][j];
					} else {
						match[i % 2][j] = match[i % 2][j - 1];
					}
				}
			}
		}

		printf("%d\n", match[(i - 1) % 2][j - 1]);
	}
}