Common Subsequence（HDU-1159）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37077    Accepted Submission(s): 16977

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
  
  

 

题目大意：

给你两串字符串，求出最长的相同序列，此序列不一定相邻，如所给例子一，最长相同序列为 abfc。

解题思路：

DP的类型题，状态转移方程很巧妙，给当前的两个字符串都加一个字符：

①如果两个字符不一样，那么对原来的状态没有改变，最长长度为备选的两个 ”最长子序列 “中较长的那个。

②如果两个字符都一样，那么“最长子序列”是之前的“最长子序列”长度+1。

即：

dp[i][j]=dp[i-1][j-1]+1；（a[i]==b[j]）

dp[i][j]=max(dp[i-1][j],dp[i][j-1])(a[i]!=b[j]);

（PS：把两个数组分别作为行列建立二维数组可能会更直观一些，最后二维数组最右下角的值一定是最大值！）

代码如下：

#include<cstdio>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
string a,b;
int dp[1001][1001];
int DP();
int main()
{
	while(cin>>a>>b)
	{
		memset(dp,0,sizeof(dp));
		printf("%d\n",DP());
	}
	return 0;
}
int DP()
{
	int hang = a.length();
	int lie = b.length();
	for(int i=0 ; i<hang ; i++)
	{
		for(int j=0 ; j<lie ; j++)
		{
			if(a[i]==b[j]) dp[i+1][j+1] = dp[i][j]+1;
			else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
		}
	}
	return dp[hang][lie];
}

题目传送门。。。。。