7-2 Train scheduling (25 points)

Topic:

Sample input:

9
8 4 2 5 3 9 1 6 7

Sample output:

4

Ideas:

To get the fewest scheduling sequences, find the minimum number of descent sequences. Take the above example: there are four descending sequences

8 4 2 1

5 3

9 6

7

So only four dispatch queues are required.

Also according to the theorem: the minimum number of descent sequences equals the length of the longest ascending subsequence. (This theorem proves not understand, direct Meng, dish is original sin ah!!) The rest is a bare longest ascending subsequence problem.

Code:

#include <bits/stdc++.h>#defineINF 0x3f3f3f3fusing namespaceStd;typedefLong Longll;Const intMAXN = 1e5+Ten;intA[MAXN],DP[MAXN];intMain () {intN; scanf ("%d",&N);  for(inti =0; i<n; i++) {scanf ("%d",&A[i]); Dp[i]=inf; }    intMmax =-1;  for(inti =0; i<n; i++)    {        intK = Lower_bound (Dp,dp+n,a[i])-DP; DP[K]=A[i]; Mmax= Max (Mmax, K +1); } printf ("%d\n", Mmax); return 0;}

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7-2 Train scheduling (25 points)