UVa 111 - History Grading

111 - History Grading

 * http://acm.uva.es/p/v1/111.html

Summary
This is a DP Problem - a modified Longest Increasing Subsequence. Also it can be solved as Longest Common Subsequence but LIS is more clear.

Explanation
To find Longest Common Subsequence you can use DP: Let we have 2 sequences a[1..n], b[1..n]. And table m[0..n,0..n] were m[i,j] if the longest common subsequence of a[1..i] and b[1..j]. Then for each (i,j):

$$m[i,j]=m[i-1,j-1]+1, a[i]=b[j]$$

$$m[i,j]=max(m[i-1,j],m[i,j-1]), a[i]<>b[j]$$

Gotchas
In problem you are given not exactly comparable sequences. Input "5 2 3 1 4" means that 1 must go 5th, 2 must go 2d, 3 must go 3d, 4 must go 1st and 5 must go 4th.

Input1
4 4 2 3 1 1 3 2 4 3 2 1 4 2 3 4 1

Output1
1 2 3

Input2
10 3 1 2 4 9 5 10 6 8 7 1 2 3 4 5 6 7 8 9 10 4 7 2 3 10 6 9 1 5 8 3 1 2 4 9 5 10 6 8 7 2 10 1 3 8 4 9 5 7 6

Output2
6 5 10 9

Solution
C++: http://acm-solution.blogspot.com/2011/09/acm-uva-111-history-grading.html