FZU Problem 1160 Common Subsequence

Problem 1160 Common Subsequence

Accept: 484 Submit: 1027 Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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, 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.

Input

The program input is from a text file. Each data set in the file 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

SEERC 2003

解决方案

LCS动态规划

理论：https://blog.youkuaiyun.com/someone_and_anyone/article/details/81044153

本题不要求输出LCS序列，故只要懂得LCS的状态转移方程即可

#include<iostream>
#include<algorithm>
using namespace std;
char s1[1005], s2[1005];
int l1, l2, dp[1005][1005];
int main() {
	while (~scanf("%s%s", s1 + 1, s2 + 1)) {
		l1 = strlen(s1 + 1);
		l2 = strlen(s2 + 1);
		for (int i = 0; i <= l2; i++)
			dp[0][i] = 0;
		for (int i = 0; i <= l1; i++)
			dp[i][0] = 0;
		for (int i = 1; i <= l1; i++)
			for (int j = 1; j <= l2; j++) {
				if (s1[i] == s2[j])
					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[l1][l2]);
	}
	return 0;
}