最长公共子序列（动态规划）

参考： 

https://blog.youkuaiyun.com/hrn1216/article/details/51534607 

 

 

（S1={1,3,4,5,6,7,7,8}和S2={3,5,7,4,8,6,7,8,2}），并结合上图来说：
       假如S1的最后一个元素 与 S2的最后一个元素相等，那么S1和S2的LCS就等于 {S1减去最后一个元素} 与 {S2减去最后一个元素} 的 LCS  再加上 S1和S2相等的最后一个元素。
       假如S1的最后一个元素 与 S2的最后一个元素不等（本例子就是属于这种情况），那么S1和S2的LCS就等于 ： {S1减去最后一个元素} 与 S2 的LCS， {S2减去最后一个元素} 与 S1 的LCS 中的最大的那个序列。

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51684    Accepted Submission(s): 23804

 

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：给出两个字符串，求两个字符串的最长公共字串。

#include<bits/stdc++.h>


using namespace std;

int dp[1005][1005];

int main()
{

	string s1,s2;

	while(cin>>s1>>s2)
    {

		int len1=s1.length();

		int len2=s2.length();

		memset(dp,0,sizeof(dp));

		for(int i=1;i<=len1;i++)
                {

			for(int j=1;j<=len2;j++)
                        {

				if(s1[i-1]==s2[j-1])  dp[i][j]=dp[i-1][j-1]+1;  //状态 +1

				else  dp[i][j]=max(dp[i][j-1],dp[i-1][j]);     //状态转移

			}

		}

		printf("%d\n",dp[len1][len2]);
    }

	return 0;

}

利用滚动数组的空间优化，pre[]记录i-1的状态，dp[]现在状态。

#include<bits/stdc++.h>
using namespace std;
int dp[1005],pre[1005];
int main(){

	string s1,s2;

	while(cin>>s1>>s2){

		int len1=s1.length();

		int len2=s2.length();

		memset(dp,0,sizeof(dp));

		memset(pre,0,sizeof(pre));

		for(int i=1;i<=len1;i++)
        {

			for(int j=1;j<=len2;j++)
            {

				if(s1[i-1]==s2[j-1])  dp[j]=pre[j-1]+1;

				else  dp[j]=max(dp[j-1],pre[j]);

				pre[j-1]=dp[j-1];  //pre记录第i-1时行最长长度，dp记录第i行时最长长度
            }
			pre[len2]=dp[len2];
        }
        printf("%d\n",dp[len2]);

}
	return 0;

}