LCS 最大公共序列算法

LCS：给出两个序列S1和S2，求出的这两个序列的最大公共部分S3就是就是S1和S2的最长公共子序列了。公共部分

必须是以相同的顺序出现，但是不必要是连续的。

1、字符相同，则指向左上，并加1

2、字符不同，则指向左边或者上边较大的那个

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43195		Accepted: 17515

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, x _ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

Source

Southeastern Europe 2003

 

#include<stdio.h>
#include<string.h>
char a[1010],b[1010];
int p[1010][1010];
int max(int x,int y)
{
	return x>y?x:y;
}
int main()
{
	int i,j,str1,str2;
	while(~scanf("%s%s",a,b))
	{
		memset(p,0,sizeof(p));
		str1=strlen(a);
		str2=strlen(b);
		for(i=1;i<=str1;i++)
		{
			for(j=1;j<=str2;j++)
			{
				if(a[i-1]==b[j-1])
				{
					p[i][j]=p[i-1][j-1]+1;//字符相同，则指向左上，并加1

				}
				else {
				p[i][j]=max(p[i-1][j],p[i][j-1]);//字符不同，比较左边和上边那个大取最大 
				}
			}
		}
		printf("%d\n",p[str1][str2]);
	}
	return 0;
}