hdu 1159 Common Subsequence(DP最长公共子序列)

最长公共子序列算法实现

最新推荐文章于 2021-09-14 16:12:42 发布

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本文介绍了如何使用动态规划解决最长公共子序列问题，并通过实例展示了算法的应用。

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14714 Accepted Submission(s): 6106

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

Source

Southeastern Europe 2003

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Ignatius

思路：dp[x][y],串a前x个和串b前y个的最长公共子序列。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=5000;
char a[mm],b[mm];
int dp[mm][mm];
int lena,lenb;
int main()
{
  while(cin>>a>>b)
  {
    lena=strlen(a);lenb=strlen(b);
    for(int i=0;i<lena;i++)dp[i][0]=0;
    for(int i=0;i<lenb;i++)dp[0][i]=0;
    for(int i=1;i<=lena;i++)
      for(int j=1;j<=lenb;j++)
      if(a[i-1]==b[j-1])dp[i][j]=dp[i-1][j-1]+1;
      else dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1];
      cout<<dp[lena][lenb]<<"\n";
  }
}