UVA 10869-brownie Points II (tree array)

UVA 10869-brownie Points II

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Test instructions: N points on a plane, two persons, first select a line through the vertical x-axis of the point. And then there's a guy who crosses the line on this wire. Select a line that is perpendicular to the line, then divide the 4 quadrants, and the first person gets a score of 1. 3 quadrants, and the second person is 24 quadrants. Ask the first person according to the optimal way, can get the maximum value of the minimum score, and this value there is a person's score possible situation

Idea: Tree-like array, can enumerate a point, suppose can find the right upper and lower left point number is good, its useful a tree-like array, the y-coordinate discretization, and then recorded in, and then put the point x from left to right, each time after the deletion of points to query the current height of the number of points, so that you can get the number of upper right corner, The same can be asked to the lower left corner, can handle this step is good to run. Finally pay attention to the answer to take weight and increment, with set on OK

Code:

#include <cstdio> #include <cstring> #include <algorithm> #include <set>using namespace std;#    Define Lowbit (x) (x& (×)) const int N = 200005;struct point {int x, y, rank, ans, Mi; int xn, yn;}    P[n];int bit[n];set<int> out;void Add (int x, int v) {while (X < N) {Bit[x] + = v;x + = lowbit (x);    }}int get (int x) {int ans = 0;    while (x > 0) {ans + = bit[x];x-= lowbit (x); } return ans; int get (int l, int r) {return get (R)-Get (L-1);} BOOL Cmpy (Point A, point B) {return a.y < b.y;}    BOOL Cmpx (Point A, point B) {if (a.x = = b.x) return a.y > B.y; return a.x < b.x;}    BOOL Cmpx2 (Point A, point B) {if (a.x = = b.x) return A.Y < b.y; return a.x > b.x;}    int N, top;void init () {memset (bit, 0, sizeof (bit));    for (int i = 0; i < n; i++) {scanf ("%d%d", &p[i].x, &p[i].y);p [I].ans = 0;p[i].xn = P[i].yn = 0; } sort (p, p + N, cmpy);}    void Solve () {top = 1; for (int i = 0; i < n; i++) {iF (i && p[i].y! = p[i-1].y) top++;p [I].rank = Top;add (P[i].rank, 1);  } for (int i = 0; i < n; i++) {int J;int len = 0;for (j = i; p[i].y = = P[j].y && J < N; j + +) Len++;for    (j = i; p[i].y = = P[j].y && J < N; j + +) P[j].yn = Len;i = j-1;    } sort (p, p + N, cmpx);    for (int i = 0; i < n; i++) {Add (P[i].rank,-1);p [I].ans + = Get (P[i].rank + 1, top);  } for (int i = 0; i < n; i++) {int J;int len = 0;for (j = i; p[i].x = = p[j].x && J < N; j + +) Len++;for    (j = i; p[i].x = = p[j].x && J < N; j + +) P[j].xn = Len;i = j-1;    } memset (bit, 0, sizeof (bit));    for (int i = 0; i < n; i++) Add (P[i].rank, 1);    Sort (p, p + N, cmpx2);    for (int i = 0; i < n; i++) {Add (P[i].rank,-1);p [I].ans + = Get (1, p[i].rank-1);    } for (int i = 0; i < n; i++) {int J;int Min = 1000000000;for (j = i; p[i].x = = p[j].x && J < N; j + +) min = min (min, P[j].ans); for (j = i; p[i].x = = p[j].x && J < N; J + +) P[j].    Mi = Min;i = j-1;    } int Max = 0;    Out.clear (); for (int i = 0; i < n; i++) {if (P[i].    Mi! = P[i].ans) continue;if (P[i].ans > Max) {max = P[i].ans;    Out.clear (); Out.insert (N-p[i].ans-p[i].xn-p[i].yn + 1);}    else if (P[i].ans = = Max) {Out.insert (N-p[i].ans-p[i].xn-p[i].yn + 1);} } printf ("Stan:%d;    Ollie: ", Max);    for (Set<int>::iterator it = Out.begin (); It! = Out.end (); it++) printf ("%d", *it); printf ("; \ n");}    int main () {while (~SCANF ("%d", &n) && N) {init (); solve (); } return 0;}

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UVA 10869-brownie Points II (tree array)