UVA 10405 Longest Common Subsequence (dp + LCS)

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
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

题意：给定两个序列，求最长公共子序列。

思路：dp中的LCS问题。。裸的很水。状态转移方程为

字符相同时: d[i][j] = d[i - 1][j - 1] + 1,不同时:d[i][j] = max(d[i - 1][j], d[i][j - 1])

代码：

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

char a[1005], b[1005];
int d[1005][1005], i, j;

int max(int a, int b) {
	return a > b ? a : b;
}
int main() {
	while (gets(a) != NULL) {
		gets(b);
		memset(d, 0, sizeof(d));
		int lena = strlen(a);
		int lenb = strlen(b);
		for (i  = 1; i <= lena; i ++)
			for (j = 1; j <= lenb; j ++) {
				if (a[i - 1] == b[j - 1]) {
					d[i][j] = d[i - 1][j - 1] + 1;
				}
				else {
					d[i][j] = max(d[i - 1][j], d[i][j - 1]);
				}
			}
		printf("%d
", d[lena][lenb]);
	}
	return 0;
}