动态规划解最长公共子序列 - Common Subsequence - HDU - 1159

Common Subsequence

传送门

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.

翻译

给定序列的子序列是给定序列，其中省略了一些元素（可能没有）。给定一个序列 X = <x1, x2, …, xm> 另一个序列 Z = <z1, z2, …, zk> 是 X 的子序列，如果存在严格递增的序列 <i1, i2, … ., ik> X 的索引使得对于所有 j = 1,2,…,k, xij = zj。例如，Z = <a, b, f, c> 是 X = <a, b, c, f, b, c> 的子序列，索引序列为 <1, 2, 4, 6>。给定两个序列 X 和 Y，问题是找到 X 和 Y 的最大长度公共子序列的长度。
程序输入来自文本文件。文件中的每个数据集都包含代表给定序列的两个字符串。序列由任意数量的空格分隔。输入数据正确。对于每组数据，程序在标准输出上打印从单独行开始的最大长度公共子序列的长度。

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

分析

先是对两个字符串进行循环 当遇到相同字母时 更新dp[i+1][j+1]=dp[i][j]+1此时公共最大字符数为dp[i][j]+1
若当前的两个字符串所对应的字母不同时则其在dp[i][j+1],dp[i+1][j]中选择最大者赋值至dp[i+1][j+1]

相同	不同
dp[i][j]+1	max(dp[i][j+1],dp[i+1][j])

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1010][1010];
char a[1010],b[1010];
int main()
{
	int la,lb,i,j;
	while(~scanf("%s %s",a,b))
	{
		la=strlen(a);
		lb=strlen(b);
		memset(dp,0,sizeof(dp));
		for(i=0; i<la; i++)
			for(j=0; j<lb; j++)
			{
				if(a[i]==b[j]) dp[i+1][j+1]=dp[i][j]+1;
				else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
			}
		printf("%d\n",dp[la][lb]);
	}
}