POJ 1458 Common Subsequence

题目

总时间限制: 1000ms 内存限制: 65536kB

描述

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.

输入

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.

输出

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.

样例输入

abcfbc abfcab

programming contest

abcd mnp

样例输出

4

2

0

来源

Southeastern Europe 2003

思路

maxlen(i,j)表示s1i和s2j最长公共子序列的长度。

maxlen(i,j)={maxlen(i−1,j−1)+1, if s1i==s2jMax{maxlen(i−1,j),maxlen(i,j−1)},s1i≠s2j

代码

while True:
try:
s1, s2 = input().strip().split()
l1 = len(s1)
l2 = len(s2)
dp = [[0 for i in range(l2 + 1)] for j in range(l1 + 1)]
for i in range(l1):
for j in range(l2):
if s1[i] == s2[j]:
dp[i + 1][j + 1] = dp[i][j] + 1
else:
dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j])
print(dp[l1][l2])
except:
break