[TJOI2013]松鼠聚会

坐标变换

看图

B - G 是原来的一个单位

B - J 也是。

我们发现 切比雪夫距离 可以在坐标旋转后转换为 曼哈顿距离 。

而曼哈顿距离可以在O(n)内计算出(求出每个x到xi的距离，同理求y，相加求和即可);

具体可以看我的代码:

/**
 * Problem:Meeting
 * Author:Shun Yao
 * Time:2013.5.21
 * Result:Accepted
 */

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>

long n;
long long sumx, sumy, ans[100000], ANS;

class anode {
public:
	long long x, y;
	long id;
	anode() {}
	~anode() {}
} a[100000];

char cmpx(anode a, anode b) {
	return a.x < b.x;
}

char cmpy(anode a, anode b) {
	return a.y < b.y;
}

int main() {
	static long i;
	static long long xx, yy, sum;
	freopen("meeting.in", "r", stdin);
	freopen("meeting.out", "w", stdout);
	scanf("%ld", &n);
	sumx = sumy = 0;
	for (i = 0; i < n; ++i) {
		scanf("%lld%lld", &a[i].x, &a[i].y);
		sumx += a[i].x += a[i].y;
		sumy += a[i].y = a[i].x - (a[i].y << 1);
		a[i].id = i;
	}
	ANS = 1LL << 60;
	std::sort(a, a + n, cmpx);
	sum = 0;
	for (i = 0; i < n; ++i) {
		sum += a[i].x;
		ans[a[i].id] = ((i + 1 << 1) - n) * a[i].x + sumx - (sum << 1);
	}
	std::sort(a, a + n, cmpy);
	sum = 0;
	for (i = 0; i < n; ++i) {
		sum += a[i].y;
		ans[a[i].id] += ((i + 1 << 1) - n) * a[i].y + sumy - (sum << 1);
		if (ANS > ans[a[i].id])
			ANS = ans[a[i].id];
	}
	printf("%lld", ANS >> 1);
	fclose(stdin);
	fclose(stdout);
	return 0;
}