Common Subsequence 最长公共子序列 hdu 1159

                   Common Subsequence

           Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
           Total Submission(s): 17709 Accepted Submission(s): 7445

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
   
   
   
   

Source

Southeastern Europe 2003

Recommend

Ignatius

最长公共子序列，想了老半天都没想出状态方程~ 

它的原理是这样的~

最长公共子序列问题具有最优子结构性质

设
X = { x1 , ... , xm }
Y = { y1 , ... , yn }
及它们的最长子序列
Z = { z1 , ... , zk }

则

1、若 xm = yn ， 则 zk = xm = yn，且Z[k-1] 是 X[m-1] 和 Y[n-1] 的最长公共子序列
2、若 xm != yn ，且 zk != xm , 则 Z 是 X[m-1] 和 Y 的最长公共子序列
3、若 xm != yn , 且 zk != yn , 则 Z 是 Y[n-1] 和 X 的最长公共子序列

由性质导出子问题的递归结构

当 i = 0 , j = 0 时 , c[i][j] = 0
当 i , j > 0 ; xi = yi 时 , c[i][j] = c[i-1][j-1] + 1
当 i , j > 0 ; xi != yi 时 , c[i][j] = max { c[i][j-1] , c[i-1][j] }

注意边界，两个串的下标都是从1开始的，因为递归时是以0为结束标志的~

dp[i][j]  表示的是 表示长度为i的字符串和长度为j的字符串的最大公共子串的长度

code:

#include<iostream>
#include<string.h>
using namespace std;
char str1[1005];
char str2[1005];
int dp[1005][1005];
int main()
{
	while(scanf("%s %s",str1+1,str2+1)!=EOF)
	{	
	        memset(dp,0,sizeof(dp));这一步就已经初始化边界了~ dp[0][j:1->len2]=0;
	//	memset(str1,0,sizeof(str1));//脑袋抽抽了 刚输进去就给清了~ 
	//	memset(str2,0,sizeof(str2));
		
		int len1=strlen(str1+1),len2=strlen(str2+1);
	
	    for(int i=1;i<=len1;i++)
	     for(int j=1;j<=len2;j++)
	     {
     		if(str1[i]==str2[j])
     		dp[i][j]=dp[i-1][j-1]+1;
     		else
     		dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
     		
     	}
     	cout<<dp[len1][len2]<<endl; 
		
	}
	return 0;
}