bit hit pan ... represent 否則 pro example ber

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\(Y = (y_1,y_2,y_3,...)\)的最長公共子序列\(LCS(X,Y)\)
• 最優子結構證明
  • 如果\(x_n==y_n\)那麽\(LCS(X_{n-1},Y_{n-1})\)就是一個子問題
  • 否則就是$，，LCS = max(LCS(X_{n-1},Y_m)，LCS(X_n，Y_{m-1})) $

證明完成之後，顯然可以容易得到狀態轉移式，用\(f[i][j]\)

表示長度為i的s1和長度為j的s2的最長公共子序列的長度，詳見註釋：

#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));//初始歸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，就是比較前n-1長度的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劃分到子問題裏
        cout << f[len1][len2] << endl;
    }
    return 0;
}

Hdoj 1159.Common Subsequence 題解