HDU 1159:(DP,最长公共子序列LCS)

本文解析了一个经典的算法问题——求解两个字符串的最长公共子序列,并提供了详细的AC代码实现。通过动态规划的方法,有效地解决了该问题。

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159

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


题意:

   求两个字符串里的最大公共子序列

   什么你不知道什么是最大公共子序列,先看下好的博客吧:博客链接

思路:

  别人的博客里已经讲的很清楚了,毕竟这是个模板题,我作为一个萌新,只能默默瞻仰了,算法好高深啊,我现在学的可能只是九牛一毛吧......何时才能领悟到算法的核心呢,是一生的证明之路吗,ε=(´ο`*)))唉

AC代码:

#include <cstring>
#include <iostream>
#include <cstdio>
using namespace std;
int dp[1005][1005];
int main()
{
    string a,b;
    while( cin >> a >> b )
    {
        memset(dp,0,sizeof(dp));
        for( int i=1; i<=a.length(); i++ )
        {
            for( int j=1; j<=b.length(); 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[a.length()][b.length()]);
    }
    return 0;
}


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