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Describe ?an? algorithm ?that ?takes ?an ?unsorted ?array ?of ?axis‐aligned ?rectangles ?and? returns ?any ?pair ?of ?rectangles ?that? overlaps,? if ?there? is ?such? a ?pair.? ?Axis‐aligned? means? that? all ?the? rectangle? sides? are ?either ?parallel? or? perpendicular ?to? the? x‐? and? y‐axis. ??You? can? assume ?that? each? rectangle? object? has? two ?variables?in? it: ?the ?x‐y? coordinates ?of ?the? upper‐left ?corner ?and ?the ?bottom‐right ?corner.
Good? Answer: ?Create? a ?sorted ?array ?of? the? x? coordinates ?of ?the? left? and? right? edges ?of? the ?rectangles. ??Then, ?use ?a? "scanline" ?to? move? from? left ?to ?right ?through? the? rectangles. ??Keep ?a? binary? search? tree ?containing ?the? y ?coordinates ?of ?the? top? and? bottom? edges ?of ?the? rectangles? that? overlap ?the ?scanline. ??For ?each? element? of ?the? array,? check? whether ?it ?is ?a? left ?or ?right? edge.? ?If ?it ?is ?a ?right? edge, ?remove? the? corresponding? top? and ?bottom? edges ?from ?the ?BST.?? If ?it ?is ?a?left ?edge,? search? the? BST? for ?rectangles? that? overlap ?the? current? rectangle; ?if ?there? is ?one,?return ?the? overlap.?? Then,? add ?the?y? coordinates ?of? the? top? and? bottom?edges? of ?the ?rectangle? to ?the ?BST.?? The? search? takes ?O(n?log?n) ?time,? since ?it ?takes ?O(n?log?n) ?time? to? sort? the? rectangles? and? each? of ?the? 2n ?iteration s?takes ?O(log?n)? time.
C++ Sample Code (Find all pairs of rectangles that overlap):
1 #include <iostream> 2 #include <vector> 3 #include <map> 4 #include <algorithm> 5 using namespace std; 6 7 struct rect { 8 int _x1, _x2, _y1, _y2; 9 rect(int x1, int x2, int y1, int y2) : 10 _x1(x1), _x2(x2), _y1(y1), _y2(y2) {} 11 }; 12 13 struct xedge { 14 int _x; 15 int _is_left; 16 int _i_rect; 17 xedge(int x, int is_left, int i_rect) : 18 _x(x), _is_left(is_left), _i_rect(i_rect) {} 19 }; 20 21 struct yedge { 22 int _y; 23 int _is_top; 24 int _i_rect; 25 yedge(int y, int is_top, int i_rect) : 26 _y(y), _is_top(is_top), _i_rect(i_rect) {} 27 }; 28 29 int main() { 30 vector<rect> rcts; // vector that stores rectangles 31 vector<xedge> xes; // left and right edges of the rectangles 32 int x1, x2, y1, y2; 33 int i = 0; 34 while (cin >> x1 >> x2 >> y1 >> y2) { 35 rcts.push_back(rect(x1, x2, y1, y2)); 36 xes.push_back(xedge(x1, 1, i)); 37 xes.push_back(xedge(x2, 0, i)); 38 i++; 39 } 40 41 // sort left and right rectangles according to their x-coordinates 42 sort(xes.begin(), xes.end(), 43 [](const xedge& e1, const xedge& e2) { 44 if (e1._x != e2._x) return e1._x < e2._x; 45 else if (e1._is_left != e2._is_left) 46 return e1._is_left > e2._is_left; 47 else return e1._i_rect < e2._i_rect; 48 }); 49 50 // a binary search tree for top and bottom edges that overlap the "scanline" 51 multimap<int, yedge> edge_map; 52 for (auto& xe : xes) { // xe : "scanline" 53 if (!xe._is_left) { 54 auto range = edge_map.equal_range(rcts[xe._i_rect]._y1); 55 for (auto iter = range.first; iter != range.second; ++iter) { 56 if (iter->second._i_rect == xe._i_rect) { 57 edge_map.erase(iter); 58 break; 59 } 60 } 61 range = edge_map.equal_range(rcts[xe._i_rect]._y2); 62 for (auto iter = range.first; iter != range.second; ++iter) { 63 if (iter->second._i_rect == xe._i_rect) { 64 edge_map.erase(iter); 65 break; 66 } 67 } 68 } else { 69 int y1 = rcts[xe._i_rect]._y1; 70 int y2 = rcts[xe._i_rect]._y2; 71 auto iter1 = edge_map.lower_bound(y1); 72 auto iter2 = edge_map.upper_bound(y2); 73 for (auto iter = iter1; iter != iter2; ++iter) { 74 cout << xe._i_rect << " " << iter->second._i_rect << endl; 75 } 76 edge_map.insert(make_pair(y1, yedge(y1, 0, xe._i_rect))); 77 edge_map.insert(make_pair(y2, yedge(y2, 1, xe._i_rect))); 78 } 79 } 80 81 return 0; 82 }
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原文地址：http://www.cnblogs.com/bestofme/p/4603359.html
