Predpogoji: BIT Glede na 'n' odsekov črte je vsak od njih vodoraven ali navpičen, poiščite največje število trikotnikov (vključno s trikotniki z ničelno ploščino), ki jih je mogoče oblikovati s spajanjem presečišč odsekov črte. Nobena dva vodoravna črtna segmenta se ne prekrivata niti dva navpična črtna segmenta. Premica je predstavljena z dvema točkama (štiri cela števila, od katerih sta prvi dve koordinati x in y za prvo točko, drugi dve pa koordinati x in y za drugo točko) Primeri:
| ---|-------|-- | | ----- | --|--|- | | | | For the above line segments there are four points of intersection between vertical and horizontal lines every three out of which form a triangle so there can be 4C3 triangles.
Ideja temelji na Algoritem črtne črte . Gradnja rešitve po korakih:
- Shranite obe točki vseh odsekov črte z ustreznim dogodkom (opisanim spodaj) v vektor in razvrstite vse točke v nepadajočem vrstnem redu njihovih koordinat x.
- Zdaj si predstavljajmo navpično črto, s katero potegnemo čez vse te točke in opišemo 3 dogodke glede na to, na kateri točki trenutno smo:
- a navpična črta
- Pokličemo regijo 'aktiven' ali vodoravne črte 'aktiven' ki so imeli prvi dogodek, drugega pa ne. Imeli bomo BIT (Binarno indeksirano drevo) za shranjevanje koordinat 'y' vseh aktivnih linij.
- Ko linija postane neaktivna, odstranimo njen 'y' iz BIT-a.
- Ko pride do dogodka tretje vrste, tj. ko smo na navpični črti, poizvedujemo po drevesu v območju njegovih koordinat 'y' in rezultat prištejemo številu dosedanjih presečišč.
- Končno bomo povedali število točk presečišč m potem bo število trikotnikov (vključno z ničelno površino). mC3 .
v - skrajna leva točka odseka vodoravne črteven - skrajna desna točka odseka vodoravne črteOpomba: Moramo skrbno razvrstiti točke pogled na cmp() funkcijo v izvedbi za pojasnilo.
CPP// A C++ implementation of the above idea #include
#define maxy 1000005 #define maxn 10005 using namespace std; // structure to store point struct point { int x y; point(int a int b) { x = a y = b; } }; // Note: Global arrays are initially zero // array to store BIT and vector to store // the points and their corresponding event number // in the second field of the pair int bit[maxy]; vector<pair<point int> > events; // compare function to sort in order of non-decreasing // x coordinate and if x coordinates are same then // order on the basis of events on the points bool cmp(pair<point int> &a pair<point int> &b) { if ( a.first.x != b.first.x ) return a.first.x < b.first.x; //if the x coordinates are same else { // both points are of the same vertical line if (a.second == 3 && b.second == 3) { return true; } // if an 'in' event occurs before 'vertical' // line event for the same x coordinate else if (a.second == 1 && b.second == 3) { return true; } // if a 'vertical' line comes before an 'in' // event for the same x coordinate swap them else if (a.second == 3 && b.second == 1) { return false; } // if an 'out' event occurs before a 'vertical' // line event for the same x coordinate swap. else if (a.second == 2 && b.second == 3) { return false; } //in all other situations return true; } } // update(y 1) inserts a horizontal line at y coordinate // in an active region while update(y -1) removes it void update(int idx int val) { while (idx < maxn) { bit[idx] += val; idx += idx & (-idx); } } // returns the number of lines in active region whose y // coordinate is between 1 and idx int query(int idx) { int res = 0; while (idx > 0) { res += bit[idx]; idx -= idx & (-idx); } return res; } // inserts a line segment void insertLine(point a point b) { // if it is a horizontal line if (a.y == b.y) { int beg = min(a.x b.x); int end = max(a.x b.x); // the second field in the pair is the event number events.push_back(make_pair(point(beg a.y) 1)); events.push_back(make_pair(point(end a.y) 2)); } //if it is a vertical line else { int up = max(b.y a.y); int low = min(b.y a.y); //the second field of the pair is the event number events.push_back(make_pair(point(a.x up) 3)); events.push_back(make_pair(point(a.x low) 3)); } } // returns the number of intersection points between all // the lines vertical and horizontal to be run after the // points have been sorted using the cmp() function int findIntersectionPoints() { int intersection_pts = 0; for (int i = 0 ; i < events.size() ; i++) { //if the current point is on an 'in' event if (events[i].second == 1) { //insert the 'y' coordinate in the active region update(events[i].first.y 1); } // if current point is on an 'out' event else if (events[i].second == 2) { // remove the 'y' coordinate from the active region update(events[i].first.y -1); } // if the current point is on a 'vertical' line else { // find the range to be queried int low = events[i++].first.y; int up = events[i].first.y; intersection_pts += query(up) - query(low); } } return intersection_pts; } // returns (intersection_pts)C3 int findNumberOfTriangles() { int pts = findIntersectionPoints(); if ( pts >= 3 ) return ( pts * (pts - 1) * (pts - 2) ) / 6; else return 0; } // driver code int main() { insertLine(point(2 1) point(2 9)); insertLine(point(1 7) point(6 7)); insertLine(point(5 2) point(5 8)); insertLine(point(3 4) point(6 4)); insertLine(point(4 3) point(4 5)); insertLine(point(7 6) point(9 6)); insertLine(point(8 2) point(8 5)); // sort the points based on x coordinate // and event they are on sort(events.begin() events.end() cmp); cout << "Number of triangles are: " << findNumberOfTriangles() << "n"; return 0; } Izhod:
Number of triangles are: 4
Time Complexity: O( n * log(n) + n * log(maximum_y) )
Pomožni prostor: O(maxy), kjer je maxy = 1000005