題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

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
程序原理如下狀態方程：

#include<stdio.h> #include<string.h> #include<iostream> using namespace std;int smax(int x,int y){return x >=y ? x : y; }int maxstr[20001][20001]; char str1[2000],str2[2000];int main(){int len1,len2;while(scanf("%s%s",&str1, &str2)!=EOF){len1=strlen(str1);len2=strlen(str2);for(int i=0; i<=len1; ++i){maxstr[i][0]=0;//初始化邊界，過濾掉0的情況}for(int i=0; i<=len2; ++i){maxstr[0][i]=0;}//填充矩陣for(int i=1; i<=len1; ++i){for(int j=1; j<=len2; ++j){if(str1[i-1]==str2[j-1])//相等的情況maxstr[i][j]=maxstr[i-1][j-1]+1;else if(maxstr[i-1][j] >= maxstr[i][j-1]){//比較“左邊”和“上邊“，根據其max來填充//maxstr[i][j]=smax(maxstr[i-1][j],maxstr[i][j-1]);maxstr[i][j]=maxstr[i-1][j];}else{maxstr[i][j]=maxstr[i][j-1];}}}printf("%d\n",maxstr[len1][len2]);}return 0; }

資料鏈接：http://www.cnblogs.com/huangxincheng/archive/2012/11/11/2764625.html

總結

以上是生活随笔為你收集整理的HD 1159 Common Subsequence （最长公共子序列）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。