1159Common Subsequence（最长公共子序列）

题目：Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14543    Accepted Submission(s): 6027

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

 

题目分析：

利用动态规划

数组c[i][j]表示当x,z序列长度分别为i和j时的最长公共子序列长度

有动态方程：

1.当i=0||j=0时，c[i][j]=0;

2.当x[i]=z[j]时，c[i][j]=c[i-1][j-1]+1;

3.当x[i]!=z[j]时，c[i][j]=max{c[i-1][j]  ,   c[i][j-1]};

代码：

#include<iostream>
#include<String>
using namespace std;

int c[1000][1000];    //记录x,z序列长为i,j是的公共子序列最大长度

int length(char *x,char *z)
{
	int i,j;
	int lenx=strlen(x);
	int lenz=strlen(z);
	for(i=0;i<lenx;i++)
		c[i][0]=0;
	for(j=0;j<lenz;j++)
		c[0][j]=0;

	for(i=0;i<lenx;i++)
	{
		for(j=0;j<lenz;j++)
		{
			if(x[i]==z[j])
				c[i+1][j+1]=c[i][j]+1;
			else
			{
				if(c[i][j+1]>=c[i+1][j])
					c[i+1][j+1]=c[i][j+1];
				else
					c[i+1][j+1]=c[i+1][j];
			}
		}
	}
	return c[lenx][lenz];
}
char x[1000];
char z[1000];
int main()
{
	int len;	
	while(cin>>x>>z)
	{
		len=length(x,z);
		cout<<len<<endl;
	}
	return 0;
}