Hdoj 1159.Common Subsequence 题解

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

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思路

• (X=(x_1,x_2,x_3,..))和(Y = (y_1,y_2,y_3,...))的最长公共子序列(LCS(X,Y))
• 最优子结构证明
  • 如果(x_n==y_n)那么(LCS(X_{n-1},Y_{n-1}))就是一个子问题
  • 否则就是$，，LCS = max(LCS(X_{n-1},Y_m)，LCS(X_n，Y_{m-1})) $

证明完成之后，显然可以容易得到状态转移式，用(f[i][j])表示长度为i的s1和长度为j的s2的最长公共子序列的长度，详见注释：

#include<bits/stdc++.h>
using namespace std;
int f[1001][1001];
int main()
{
	string s1;
	string s2;
	while(cin>>s1>>s2)
	{
		int len1 = s1.size();
		int len2 = s2.size();
		memset(f,0,sizeof(f));//初始归0
		for(int i=1;i<=len1;i++)
			for(int j=1;j<=len2;j++) 
				if(s1[i-1] == s2[j-1])
					f[i][j] = max(f[i][j],f[i-1][j-1] + 1);//如果末尾相等，就是比较前n-1长度的LCS+1和自己的
				else
					f[i][j] = max(f[i][j-1], f[i-1][j]);//如果末尾元素不相等就划分到子问题里
		cout << f[len1][len2] << endl;
	}
	return 0;
}