hdoj1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13147    Accepted Submission(s): 5396

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 abfcabprogramming contest abcd mnp

 

Sample Output

420

 

Source

Southeastern Europe 2003

 

Recommend

Ignatius

动态规划的一个计算两个序列的最长公共子序列的方法如下：

以两个序列 X、Y 为例子：

设有二维数组 f[i,j] 表示 X 的 i 位和 Y 的 j 位之前的最长公共子序列的长度，则有：

f[1][1] = same(1,1);

f[i,j] = max{f[i-1][j -1] + same(i,j),f[i-1,j],f[i,j-1]}

其中，same(a,b)当 X 的第 a 位与 Y 的第 b 位完全相同时为“1”，否则为“0”。

此时，f[j]中最大的数便是 X 和 Y 的最长公共子序列的长度，依据该数组回溯，便可找出最长公共子序列。

该算法的空间、时间复杂度均为O(n^2)，经过优化后，空间复杂度可为O(n)。

#include<iostream>#include<cstring>using namespace std;char a[600], b[600];int dp[600][600];int main() {    while (cin >> a >> b) {        int len1 = strlen(a);        int len2 = strlen(b);        memset(dp, 0, sizeof (dp));        for (int i = 1; i <= len1; i++)            for (int j = 1; j <= len2; j++)                if (a[i - 1] == b[j - 1])                    dp[i][j] = dp[i - 1][j - 1] + 1;                else                    dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];        cout << dp[len1][len2] << endl;    }    return 0;}