HDU 1159.Common Subsequence（DP 最长公共子序列）超详细题解

1.题目

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.
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. 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.
Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

2.题意

求两个字符串的最长公共子序列长度。

3.思路

动态规划
1.设数组dp[i][j],表示字符串a的i号位和字符串b的j号位之前的最长公共子序列长度。(不包括i号位和j号位) 则有：
(1)若a[i-1]=b[j-1],则两个字符串的最长公共子序列长度增加一位，即有dp[i][j]=dp[i-1][j-1]+1.
(2)若a[i-1]!=b[j-1],则dp[]i[j]继承dp[i-1][j]与dp[i][j-1]的较大值，即有dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])。
2.边界dp[0…len_a][0]=0,dp[0][0…len_b]=0。
3.这样dp[i][j]只与之前的状态有关，由边界出发即可得到整个dp数组，dp[len_a][len_b]即为答案。

4.代码

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 1000;
int dp[maxn][maxn];
char a[maxn],b[maxn];
int main()
{
	while (scanf("%s%s", a, b)!=EOF)
	{
		memset(dp, 0, sizeof(dp));
		int len_a = strlen(a);
		int len_b = strlen(b);
		for (int i = 0; i <= len_a; i++) 
			dp[i][0] = 0;
		for (int i = 0; i <=len_b; i++)
			dp[0][i] = 0;
		for (int i = 1 ; i <=len_a; i++)
		{
			for (int j = 1; j <= len_b; j++)
			{
				if (a[i-1] == b[j-1])
				{
					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[len_a][len_b]);
	}
	return 0;
}