UVa 10405 Longest Common Subsequence （DP&LCS）

UVa 10405 Longest Common Subsequence （DP&LCS）2014-07-11 csdn博客 synapse710405 - 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:

abcdghaedfhr

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

a1b2c3d4ezz1yy2xx3ww4vvabcdghaedfhrabcdefghijklmnopqrstuvwxyza0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0abcdefghijklmnzyxwvutsrqpoopqrstuvwxyzabcdefghijklmn

Output for the sample input

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
", dp[lena][lenb]）;}return 0;}