UVa 10405 Longest Common Subsequence (DP&LCS)

10405 - Longest Common Subsequence

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=1346

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

模板题?


完整代码:

/*0.068s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1010;

char a[maxn], b[maxn];
int dp[maxn][maxn];

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

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