forceinline void
TaskTree<TaskView,Node>::update(int i, bool l) {
if (l)
i = _leaf[i];
assert(!n_root(i));
do {
i = n_parent(i);
node[i].update(node[n_left(i)],node[n_right(i)]);
} while (!n_root(i));
}
forceinline long long int
ExtOmegaTree<TaskView>::env(int i) {
// Enter task i
leaf(i).e = tasks[i].e();
leaf(i).env =
static_cast<long long int>(c)*tasks[i].est()+tasks[i].e();
leaf(i).cenv =
static_cast<long long int>(c-ci)*tasks[i].est()+tasks[i].e();
TaskTree<TaskView,ExtOmegaNode>::update(i);
// Perform computation of node for task with minest
int met = 0;
{
long long int e = 0;
while (!n_leaf(met)) {
if (plus(node[n_right(met)].cenv,e) >
static_cast<long long int>(c-ci) * tasks[i].lct()) {
met = n_right(met);
} else {
e += node[n_right(met)].e; met = n_left(met);
}
}
}
/*
* The following idea to compute the cut in one go is taken from:
* Joseph Scott, Filtering Algorithms for Discrete Resources,
* Master Thesis, Uppsala University, 2010 (in preparation).
*/
// Now perform split from leaf met upwards
long long int a_e = node[met].e;
long long int a_env = node[met].env;
long long int b_e = 0;
while (!n_root(met)) {
if (left(met)) {
b_e += node[n_right(n_parent(met))].e;
} else {
a_env = std::max(a_env, plus(node[n_left(n_parent(met))].env,a_e));
a_e += node[n_left(n_parent(met))].e;
}
met = n_parent(met);
}
return plus(a_env,b_e);
}
forceinline bool
TaskTree<TaskView,Node>::right(int i) {
assert(!n_root(i));
// A left node has an even number
return (i & 1) == 0;
}
forceinline bool
TaskTree<TaskView,Node>::left(int i) {
assert(!n_root(i));
// A left node has an odd number
return (i & 1) != 0;
}