POJ1518、 UVA1531 - Problem Bee（几何+贪心）

Imagine a perfectly formed honeycomb, spanning the infinite Cartesian plane. It is an interlocking grid composed of congruent equilateral hexagons. One hexagon is located so that its center is at the origin and has two corners on the X-axis. A bee must be very careful about how it travels in order not to get lost in the infinite plane. To get from an arbitrary point A to another arbitrary point B, it will first head from A to the exact center of the hexagon in which A is located. Then, it will travel in a straight line to the exact center of an adjacent hexagon. It will move from center to adjacent center until it has reached the hexagon containing point B. At the destination hexagon, it will move from the center to point B. In all cases, the bee will take a path of minimal distance that obeys the rules. The figure below demonstrates one possible minimal path from point A to point B.

Input

Input will be in the form of 5 floating point numbers per line. The first number will be the length, in centimeters, of the sides of the hexagons. The next two numbers will be the x and y coordinates of point A, followed by the x and y coordinates of point B. The input will be terminated by a line containing five zeroes. Neither point A nor point B will ever be exactly on a border between hexagons.

Output

For each line of the input, output the minimum length of a path from point A to point B, to the nearest 0.001 centimeters.

Sample Input

1.0 -3.2 2.2 3.3 09 1 4 5 10.1 0.09 0 0.21 00 0 0 0 0

Sample Output

7.7375.0000.526

题意：给定蜂窝长度，起点A和起点B，蜜蜂移动要从A移动到对应的六边形中心，然后从一个六边形中心移动到另一个六边形中心，最后在从六边形中心移动到B，注意如果AB在同一中心，直接A移动到B。

思路：竖直方向和水平方向多的肯定是要移动的段数，那么问题就变成找A和B的中心，直接找点对应的X,Y坐标是第几个六边形，注意如果在边上要处理一下，最后判断周围6个是否有比它距离更近的六边形中心，确定中心后，路径就很好计算了，

代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#define max(a,b) ((a)>(b)?(a):(b))const double pi = acos(-1.0);const int d[6][2] = {{0, 2}, {0, -2}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};double l, xa, ya, xb, yb, hd, wd;int xan, yan, xbn, ybn;double dis(double x1, double y1, double x2, double y2) {    return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));}void tra(int &x, int &y, double xa, double ya) {    if (x % 2) {if (y % 2 == 0) {    y++;}    }    else {if (y % 2) {    y++;}    }    int minx = x, miny = y;    double mind = dis(x * wd, y * hd, xa, ya);    for (int i = 0; i < 6; i++) {double xx = (d[i][0] + x) * wd;double yy = (d[i][1] + y) * hd;if (mind > dis(xx, yy, xa, ya)) {    minx = d[i][0] + x;    miny = d[i][1] + y;    mind = dis(xx, yy, xa, ya);}    }    x = minx; y = miny;}void init() {    hd = sin(pi/3) * l;    wd = l + l / 2;    xan = xa / wd;    yan = ya / hd;    xbn = xb / wd;    ybn = yb / hd;    tra(xan, yan, xa, ya); tra(xbn, ybn, xb, yb);}double solve() {    if (xan == xbn && yan == ybn)return dis(xa, ya, xb, yb);    int xx = abs(xbn - xan), yy = abs(ybn - yan);    double ans = dis(xa, ya, xan * wd, yan * hd) + dis(xb, yb, xbn * wd, ybn * hd) + xx * sin(pi/3) * 2 * l;    if (yy > xx)ans += (yy - xx) * l * sqrt(3) / 2;    return ans;}int main() {    while (~scanf("%lf%lf%lf%lf%lf", &l, &xa, &ya, &xb, &yb) && l || xa || ya || xb || yb) {init();printf("%.3lf\n", solve());    }    return 0;}