Common Subsequence--poj--1458

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17815		Accepted: 6847

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, x_ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

解题思路：这是个求最长公共子序列长度的问题，首先要知道子序列不一定必须是连续的，即中间可以被其他字符分开，但它们的顺序必须是正确的。另外，最长公共子序列不一定只有一个，而我们要寻找的是其中之一就够了。

此题是典型的动态规划问题，我们先考虑输入序列S1和S2的前缀序列：

设c[i, j] = |LCS(S1[1..i], S2[1..j])|，则有c[m, n] = |LCS(S1, S2)|。

仔细观察我们可以发现递推式：

c[i, j] = c[i-1, j-1] + 1, 如果S1[i] = S2[j]

=max{c[i-1, j], c[i, j-1]}, 如果S1[i] != S2[j]

因此我们获得了该问题的最优子结构特性，即一个最优解决方案包含了子问题的最优解决方案，也就是说，如果CS=LCS(S1, S2)，则CS的任何前缀子串必是S1和S2的某个前缀的最长公共子序列。

通过分析我们还可以知道该递归包含了重复的子问题，因此我们就可以使用DP来降低复杂度。

计算最长公共子序列长度的DP算法如下：

初始化c[i, j]，0<=i<=m, 0<=j<=n

for i=1 to m do

for j=1 to n do

if S1[i]==S2[j] then

c[i, j] = LCS[S1, S2, i-1, j-1] + 1 (即c[i-1, j-1] + 1)

else if c[i-1, j] <= c[i, j-1] then

c[i, j] = c[i, j-1]

else

c[i, j] = c[i-1, j]

代码如下：

#include <iostream>

#include <string>

std::string str1;

std::string str2;

int num[256][256]; //数组开低了会出现Runtime Error

int lcs(std::string s1, std::string s2, int m, int n)

{

for(int i=1; i<=m; i++)//此处需从1开始,因为下面循环有num[i-1][j-1]是最小的

{

for(int j=1; j<=n; j++)

{

if(s1[i-1] == s2[j-1])

{

num[i][j] = num[i-1][j-1] + 1;

}

else

{

if(num[i][j-1] < num[i-1][j])

{

num[i][j] = num[i-1][j];

}

else

{

num[i][j] = num[i][j-1];

}

}

}

}

return num[m][n];

}

int main()

{

int len1, len2;

while(std::cin>>str1>>str2)

{

memset(num, 0, sizeof(num));

len1 = str1.length();

len2 = str2.length();

int out = lcs(str1,str2,len1,len2);

std::cout<<out<<std::endl;

}

system("pause");

return 0;

}