/*
* The MIT License
*
* Copyright (c) 2010, InfraDNA, Inc.
*
* Permission is hereby granted, free of charge, to any person obtaining a copy
* of this software and associated documentation files (the "Software"), to deal
* in the Software without restriction, including without limitation the rights
* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
* copies of the Software, and to permit persons to whom the Software is
* furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in
* all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
* THE SOFTWARE.
*/
package hudson.model.queue;
import java.util.Map;
import java.util.SortedMap;
import java.util.TreeMap;
import static java.lang.Math.*;
/**
* Represents a mutable q(t), a discrete value that changes over the time.
*
* <p>
* Internally represented by a set of ranges and the value of q(t) in that range,
* as a map from "starting time of a range" to "value of q(t)".
*/
final class Timeline {
// int[] is always length=1
private final TreeMap<Long, int[]> data = new TreeMap<Long, int[]>();
/**
* Obtains q(t) for the given t.
*/
private int at(long t) {
SortedMap<Long, int[]> head = data.subMap(t,Long.MAX_VALUE);
if (head.isEmpty()) return 0;
return data.get(head.firstKey())[0];
}
/**
* Finds smallest t' > t where q(t')!=q(t)
*
* If there's no such t' this method returns null.
*/
private Long next(long t) {
SortedMap<Long, int[]> x = data.tailMap(t + 1);
return x.isEmpty() ? null : x.firstKey();
}
/**
* Splits the range set at the given timestamp (if it hasn't been split yet)
*/
private void splitAt(long t) {
if (data.containsKey(t)) return; // already split at this timestamp
SortedMap<Long, int[]> head = data.headMap(t);
int v = head.isEmpty() ? 0 : data.get(head.lastKey())[0];
data.put(t, new int[]{v});
}
/**
* increases q(t) by n for t in [start,end).
*
* @return peak value of q(t) in this range as a result of addition.
*/
int insert(long start, long end, int n) {
splitAt(start);
splitAt(end);
int peak = 0;
for (Map.Entry<Long, int[]> e : data.tailMap(start).headMap(end).entrySet()) {
peak = max(peak, e.getValue()[0] += n);
}
return peak;
}
/**
* Finds a "valley" in this timeline that fits the given duration.
* <p>
* More formally, find smallest x that:
* <ul>
* <li>x >= start
* <li>q(t) <= n for all t \in [x,x+duration)
* </ul>
*
* @return null
* if no such x was found.
*/
Long fit(long start, long duration, int n) {
OUTER:
while (true) {
long t = start;
// check if 'start' satisfies the two conditions by moving t across [start,start+duration)
while ((t-start)<duration) {
if (at(t)>n) {
// value too big. what's the next t that's worth trying?
Long nxt = next(t);
if (nxt==null) return null;
start = nxt;
continue OUTER;
} else {
Long nxt = next(t);
if (nxt==null) t = Long.MAX_VALUE;
else t = nxt;
}
}
// q(t) looks good at the entire [start,start+duration)
return start;
}
}
}