UVALive - 7528 Beehive

题目链接

分析：

我们可以建一个坐标系



竖线右边＋X，左边－X，一列为一；纵方向半个六边形高度为一；

假设坐标(-2, 2)到(3,4)，那么让(-2, 2)的横坐标先变成3这样至少三步；如果纵坐标的高度差小于等于横坐标高度差，那么高度差可以在横坐标平移的的过程中弥补，否则还需要纵坐标平移。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <cmath>

using namespace std;

const int d[] = {4, 2};

int sum[200], num[200];

int init() {
sum[1] = 0;
int t = 1, cnt = 2;
while (true) {
num[cnt - 1] = sum[cnt] = t;
cnt ++;
t += d[cnt & 1];
if (cnt > 100) break;
}
for (int i = 1; i < cnt; i ++)
sum[i] += sum[i - 1];
return cnt;
}

void work(int x, int len, int &xx, int &yy) {
int f = (int)(lower_bound(sum + 1, sum + len, x) - sum - 1);
int k = x - sum[f];
int d = (f + 1) / 2;

if (k <= d || k >= num[f] - d + 1) {
if (k <= d) {
xx = 1 - f;
yy = (k - 1) * 2 + 1;
if (f % 2 == 0) yy ++;
}
else {
xx = f - 1;
int tmpk = num[f] - k;
yy = tmpk * 2 + 1;
if (f % 2 == 0) yy ++;
}
}
else {
if (k > num[f] / 2) {
int tmpk = k - num[f] / 2;
xx = tmpk - 1;
int rnk = num[f] - k - d;
yy = rnk + d * 2;
if (f % 2 == 0) yy ++;
}
else {
xx = k - num[f] / 2 - 1;
int rnk = k - d - 1;
yy = rnk + d * 2;
if (f % 2 == 0) yy ++;
}
}
}

int main() {
int cnt = init();

int a, b;
while (~scanf("%d%d", &a, &b)) {
if (a == 0 && b == 0) break;
int x1, x2, y1, y2;
work(a, cnt, x1, y1);
work(b, cnt, x2, y2);
int dx = abs(x1 - x2);
int dy = abs(y1 - y2);
if (dy > dx) printf("%d\n", (dy - dx) / 2 + dx);
else printf("%d\n", dx);
}
return 0;
}