Beehive UVALive - 7528-优快云博客

本文介绍了一种通过建立坐标系来解决六边形网格中任意两点间最短距离的方法。利用C++实现，该算法能够快速计算出任意两个六边形节点间的距离。

There is an infinite beehive like the one given in the figure. We consider two cells to be adjacent if andonly if they share a side. A path of length k from cell c0 to cell ck is a sequence of cells c0, c1, . . . , cksuch that ci and ci+1 are adjacent for all 0 ≤ i < k. The distance between cells i and j is the length ofthe shortest path from cell i to cell j

这里写图片描述

The cells of the beehive are indexed using positive integers as shown. The cells with larger distancefrom cell 1 are given larger indices. The indices of cells with the same distance from cell 1 increasesfrom left to right. Each positive integer is the index of exactly one cell.We want to know the distance of two cells whose indices are given.

Input

There are multiple test cases in the input. Each test case is a single line containing two space-separatedintegers i and j as the indices of two cells (1 ≤ i, j ≤ 104). The input terminates with a line containing‘0 0’ which should not be processed as a test case.

output

For each test case, output a single line containing the distance of the given cells.

Sample Input

8 4

11 12

365 365

0 0

Sample Output

题意：任意给你两个六边形的号问你它们之间的最短距离

思路：建立坐标系

#include <bits/stdc++.h>
using namespace std;
#define maxn 101000
int x[maxn],y[maxn];
void init()
{
       x[1] = 0,y[1] = 0;
    for(int i = 2, num = 1; num <= 11000; i++)
    {
        int px = 0 - ( i - 1);
        int py = i % 2 ? 0 : 1;
        for(int j = (i + 1) / 2; j > 0; j--)
        {
            x[++num] = px;
            y[num] = py;
            py += 2;
        }
        py--;
        px++;
        for(int j = i - 1; j >= 1; j--)
        {
            x[++num] = px++;
            y[num]  = py++;
        }
        py-= 2;;
        for(int j = i - 1; j >= 1; j--)
        {
            x[++num] = px++;
            y[num] = py--;
        }
        py--;
        px--;
        for(int j = ( i + 1)/2 - 1; j >= 1; j-- )
        {
            x[++num] = px;
            y[num] = py;
            py -= 2;
        }
    }
}
int main()
{
    int n, m;
    init();
//    for(int i = 1; i <= 20; i++)
//        printf("%d : %d %d\n",i,x[i],y[i]);
    while(~scanf("%d%d", &n, &m))
    {
        if(!n && !m)
            break;
        int dx = abs(x[n] - x[m]);
        int dy = abs(y[n] - y[m]);
        //printf("dx: %d   dy: %d\n",dx,dy);
        if(dx >= dy)
            printf("%d\n", dx);
        else printf("%d\n", dx + (dy - dx)/2);
    }
}