POJ 1458 Common Subsequence 【最长公共子序列】

解题思路：先注意到序列和串的区别，序列不需要连续，而串是需要连续的，先由样例abcfbc         abfcab画一个表格分析，用dp[i][j]储存当比较到s1[i]，s2[j]时最长公共子序列的长度

a    b    f    c    a    b

0    0    0    0    0   0    0

a  0    1     1    1    1   1    1

b  0    1     2    2    2   2    2

c  0    1     2    2    3   3    3

f  0     1     2    3    3   3   3

b 0     1    2     3   3    3   4

c  0     1    2     3   4   4    4

其中 s1 abcfbc

s2 abfcab

以s1中的a来分析，用它与s2中的字母比较，如果相同，那么dp[i][j]=dp[i-1][j-1]+1（遇到相同的字母公共子序列的长度增加1）

如果不同的话，dp[i][j]=max(dp[i-1][j],dp[i][j-1])                 (即取相邻两种情况中的最大值)

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39737		Accepted: 15977

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

#include<stdio.h>
#include<string.h>
#define maxn 10010
int dp[maxn][maxn];

int max(int a,int b)
{
	if(a>b)
	return a;
	else
	return b;
}
int main()
{
	char s1[maxn],s2[maxn];
	int len1,len2,i,j;
	while(scanf("%s %s",&s1,&s2)!=EOF)
	{
	len1=strlen(s1);
	len2=strlen(s2);

	for(i=1;i<=len1;i++)
	{
		for(j=1;j<=len2;j++)
		if(s1[i-1]==s2[j-1])
		dp[i][j]=dp[i-1][j-1]+1;
		else
		dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
	}
	printf("%d\n",dp[len1][len2]);
    }

}