UVaOJ_10405 - Longest Common Subsequence

Problem C: 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

is 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
aedfhrabcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

Problem Setter: Piotr Rudnicki

注意数组越界，字符串中可以包含空格，所以写输入模块时要注意，不然就是Wrong answer！！！

#include <stdio.h>
#include <string.h>

int fun();

char ch1[1010];
char ch2[1010];
int LCS[1010][1010];

int main()
{
	
	while(fgets(ch1, 1010, stdin) && fgets(ch2, 1010, stdin))
	{
		printf("%d\n", fun());
	}
}

int fun()
{
	int len1 = strlen(ch1) - 1;
	int len2 = strlen(ch2) - 1;
	int i, j;
	memset(LCS, 0, sizeof(LCS)); 
	for(i = 1; i <= len1; i++)
		for(j = 1; j <= len2; j++)
		{
			if(ch1[i-1] == ch2[j-1])
				LCS[i][j] = LCS[i-1][j-1] + 1;
			else if(LCS[i-1][j] >= LCS[i][j-1])
					LCS[i][j] = LCS[i-1][j];
			else
				LCS[i][j] = LCS[i][j-1];
		}
	return LCS[len1][len2];
}