1968: Common Subsequence

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本文介绍了一个经典的计算机科学问题——寻找两个序列之间的最长公共子序列，并提供了一段C语言实现的示例代码。该代码能够接收两组字符串作为输入，并输出它们之间最长公共子序列的长度。

1968: Common Subsequence

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	4096K	607	187	Standard

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_{i_j} = 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

#include<stdio.h>
#include<string.h>
char a[500],b[500];
int c[500][500];
int main()
{
int i,j;
while(scanf("%s%s",a,b)==2)
{
  int n=strlen(a),m=strlen(b);
  for(i=0;i<=n;i++) c[i][0]=0;
  for(i=0;i<=m;i++) c[0][i]=0;
  for(i=1;i<=n;i++)
   for(j=1;j<=m;j++)
   {
    if(a[i-1]==b[j-1]) c[i][j]=c[i-1][j-1]+1;
    else if(c[i-1][j]>=c[i][j-1]) c[i][j]=c[i-1][j];
    else c[i][j]=c[i][j-1];
   }
  printf("%d/n",c[n][m]);
  //for(i=0;i<=n;i++){ for(j=0;j<=m;j++) printf("%d    ",c[i][j]);printf("/n");}
}
return 0;
}