HD 1159 Common Subsequence （最长公共子序列）

题目链接：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

程序原理如下状态方程：

#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;//初始化边界，过滤掉0的情况}for(int i=0; i<=len2; ++i){maxstr[0][i]=0;}//填充矩阵for(int i=1; i<=len1; ++i){for(int j=1; j<=len2; ++j){if(str1[i-1]==str2[j-1])//相等的情况maxstr[i][j]=maxstr[i-1][j-1]+1;else if(maxstr[i-1][j] >= maxstr[i][j-1]){//比较“左边”和“上边“，根据其max来填充//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;
}

资料链接：http://www.cnblogs.com/huangxincheng/archive/2012/11/11/2764625.html

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