2016-level algorithm sixth time on machine-e.bamboo eat me a punch

Bamboo's Eating Me punch analysis

When the distance between the two points is <=d, the fist can be punched, so that the point to meet the punch condition is the least, but not 0.

Finding the nearest point-to-distance, the nearest distance can make the number of punches possible, because other combinations are not eligible except for the nearest point.

The minimum distance to find two points in a pile of points is that the violent O (n^2) calculation is horrible, and the problem can be made smaller by splitting the mind:
Divide the points on the plane into two dials, the nearest two points may only appear in: The first pile, the second pile, and the two piles 2 each one point
Decomposition
Imagine a vertical line dividing the given point set into two dials: all points are either to the left of the line or to the right. In ascending order by x-coordinate.
Solve
Divide two recursive calls, find the nearest pair in the left, and find the nearest pair in the right one at a time. Take d ' as the minimum value in both
Merge
The nearest point-to-distance d is either a recursive-found d ', or a point pair with a point on the left-hand side. It is necessary to find out if there is a distance less than d ' and one on the left at the right point. If the distance between these two points is <d ', then the two points are actually within the distance of the vertical line d '. That is, the vertical line for the axis, width of 2*d ' range, the point in this range is taken out, and in accordance with the ascending order of Y, only need to go through a bit with the range around 7 points within the distance can be, compared to choose a smaller value.

Code

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <iomanip>using namespaceStdConst intMaxx =1e5+3;intNum[maxx];structpoint{intx, y;} P[maxx];BOOLCMPX (ConstPoint&a,Constpoint& b)//sorted by x-coordinate{returna.x<b.x;}BOOLCmpy (Const int& A,Const int&AMP;B)//Sort by y-coordinate{returnP[A].Y&LT;P[B].Y;}DoubleDis (point A, point B)//Calculate two points distance{return((Double) (a.x-b.x) * (a.x-b.x) + (Double) (A.Y-B.Y) * (A.Y-B.Y));}DoubleClosestintLowintHigh//Find the nearest distance{if(Low +1= = high)//Two point situation, already recursive to the end        returnDis (P[low], P[high]);if(Low +2= = high)//Three points        returnMin (Dis (p[low], P[high]), min (Dis (p[low), P[low +1]), DIS (P[low +1], P[high]));intMid = (low + high)/2;Doubleans = min (Closest (Low, mid), closest (Mid +1, high));//Two groups recursively, go to the bottom    intI, j, cnt =0; for(i = low, I <= high; i++) {if(p[i].x >= p[mid].x-ans && p[i].x <= p[mid].x + ans) num[cnt++] = i; } sort (num, num + cnt, cmpy);//This is ordered by the y-coordinate within the straight-line ans distance     for(i =0; I < CNT; i++) {intK = i +7> CNT? Cnt:i +7; for(j = i +1; J < K; J + +) {if(P[num[j]].y-p[num[i]].y > Ans) Break;        ans = min (dis (p[num[i]], p[num[j]), ans); }    }returnAns;}intMain () {intN while(~SCANF ("%d", &n)) { for(inti =0; i<n; i++) scanf ("%d %d", &p[i].x, &AMP;P[I].Y); Sort (p, p + N, cmpx);DoubleAns = closest (0N1);        Ans = sqrt (ans); printf"%.2lf\n", ans); }}

Other than that

Strictly speaking the data is to chuck the O (n^2) violence algorithm, but your grh with sin and cos to calculate the distance, the operation time is even faster than O (NLOGN) division,,,

2016-level algorithm sixth time on machine-e.bamboo eat me a punch