UVA 10405 - Longest Common Subsequence

                                Longest Common Subsequence

Sequence 1:                

Sequence 2:                


Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:

abcdgh
aedfhr

s  adh  of length 3.

Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

题意：求最长公共子序列。

题目大意：求最长公共子序列 设d[i][j]为A1,A2...Ai和B1,B2...Bj的LCS长度，则的d[i][j]=max{d[i-1][j],d[i][j-1]}，如果A[i]=B[j]，d[i][j]=max{d[i][j],d[i-1][j-1]+1}， 
#include<stdio.h>
#include<string.h>
int dp[1005][1005];
char s1[1005],s2[1005];
int max(int x,int y)
{
	return x>y?x:y;
}
int main()
{
	int len1,len2,i,j;
	while(gets(s1)!=NULL)
	{
		gets(s2);
		len1=strlen(s1);
		len2=strlen(s2);
		for(i=0;i<=len1;i++)
			for(j=0;j<=len2;j++)
				dp[i][j]=0;
		for( i=1;i<=len1;i++)
			for(j=1;j<=len2;j++)
			{
				dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
				if(s1[i-1]==s2[j-1])
					dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
			}
		printf("%d\n",dp[len1][len2]);
	}
	return 0;
}