[HDU - 1159] Common Subsequence（动态规划&&LCS）_hdu a subsequence of a given sequence is the given-优快云博客

本文介绍了解决最长公共子序列问题的一种动态规划方法，通过样例输入输出展示了算法的具体实现过程。针对两个字符串，该算法能找出它们之间的最长公共子序列的长度。

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

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.

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.

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

思路：

dp[i][j] ：表示第一个字符串在第 i 个字符前且第二个串在第 j 个字符前可构成的最长子序列的长度

动态方程：

dp[i+1][j+1] = 0 （i=0 || j=0）
             = dp[i][j]+1 （str1[i]==str2[j]）
             = max(dp[i+1][j],dp[i][j+1]) （str1[i]!=str2[j]）

Code：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
char s1[maxn];
char s2[maxn];
int dp[1010][1010];
int main()
{
    while(~scanf("%s %s",s1,s2))
    {
        memset(dp,0,sizeof(dp));
        int a=strlen(s1);
        int b=strlen(s2);
        for(int i=0;i<a;i++)
            for(int j=0;j<b;j++)
            {
                if(s1[i]==s2[j])
                    dp[i+1][j+1]=dp[i][j]+1;
                else
                    dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
            }
        printf("%d\n",dp[a][b]);
    }
return 0;
}