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Lines Matching full:branches

24     SkTDArray<Branch> branches;  in insert()  local
25 branches.setReserve(N); in insert()
33 Branch* b = branches.push(); in insert()
38 fCount = branches.count(); in insert()
44 n->fChildren[0] = branches[0]; in insert()
46 fRoot.fBounds = branches[0].fBounds; in insert()
49 fRoot = this->bulkLoad(&branches); in insert()
64 int SkRTree::CountNodes(int branches, SkScalar aspectRatio) { in CountNodes() argument
65 if (branches == 1) { in CountNodes()
68 int numBranches = branches / kMaxChildren; in CountNodes()
69 int remainder = branches % kMaxChildren; in CountNodes()
83 for (int j = 0; j < numTiles && currentBranch < branches; ++j) { in CountNodes()
96 for (int k = 1; k < incrementBy && currentBranch < branches; ++k) { in CountNodes()
104 SkRTree::Branch SkRTree::bulkLoad(SkTDArray<Branch>* branches, int level) { in bulkLoad() argument
105 if (branches->count() == 1) { // Only one branch. It will be the root. in bulkLoad()
106 return (*branches)[0]; in bulkLoad()
109 // We might sort our branches here, but we expect Blink gives us a reasonable x,y order. in bulkLoad()
112 int numBranches = branches->count() / kMaxChildren; in bulkLoad()
113 int remainder = branches->count() % kMaxChildren; in bulkLoad()
118 // If the remainder isn't enough to fill a node, we'll add fewer nodes to other branches. in bulkLoad()
132 for (int j = 0; j < numTiles && currentBranch < branches->count(); ++j) { in bulkLoad()
146 n->fChildren[0] = (*branches)[currentBranch]; in bulkLoad()
148 b.fBounds = (*branches)[currentBranch].fBounds; in bulkLoad()
151 for (int k = 1; k < incrementBy && currentBranch < branches->count(); ++k) { in bulkLoad()
152 b.fBounds.join((*branches)[currentBranch].fBounds); in bulkLoad()
153 n->fChildren[k] = (*branches)[currentBranch]; in bulkLoad()
157 (*branches)[newBranches] = b; in bulkLoad()
161 branches->setCount(newBranches); in bulkLoad()
162 return this->bulkLoad(branches, level + 1); in bulkLoad()