[CF535E] Tavas and Pashmaks

题面

题解

我们可以尝试寻找临界值。枚举，那么令(frac{A}{a_i}+frac{B}{b_i}=frac{A}{a_j}+frac{B}{b_j})，如果这对(A,B)在(i,j)取到最值，那么(i,j)有用。

将每个型号看成平面上的点((frac1{a_i},frac1{b_i}))，我们的问题变成了：给定任意(A,B)，最小化(z=Ax+By)（(x,y)即为点的坐标）。因为(A,B)为正实数,所以目标函数的斜率为负。于是，有用的点分布在该点集的左下凸包上。使用类似斜率优化的方法求出凸包即可。

时间复杂度(O(nlogn))

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));

inline int read()
{
	int data = 0, w = 1;
	char ch = getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data*w;
}

const int maxn(3e5 + 10);
int q[maxn], top, next[maxn], ok[maxn], n;
struct point { int x, y, id; } p[maxn];
double k[maxn];
inline bool operator < (const point &lhs, const point &rhs)
{
	return lhs.x > rhs.x || (lhs.x == rhs.x && lhs.y > rhs.y);
}

inline double slope(const point &i, const point &j)
{
	return 1. * i.x * j.x * (j.y - i.y) / (1. * i.y * j.y * (j.x - i.x));
}

int main()
{
	n = read(); int minx, miny = 0;
	for(RG int i = 1; i <= n; i++)
	{
		p[i] = (point) {read(), read(), i};
		if(miny < p[i].y || (miny == p[i].y && minx < p[i].x))
			minx = p[i].x, miny = p[i].y;
	}

	std::sort(p + 1, p + n + 1);
	q[top = 1] = 1;
	for(RG int i = 2; i <= n && minx <= p[i].x; i++)
	{
		if(p[q[top]].x == p[i].x)
		{
			if(p[q[top]].y == p[i].y)
				next[p[i].id] = next[p[q[top]].id], next[p[q[top]].id] = p[i].id;
			continue;
		}

		while(top > 1 && k[top] > slope(p[q[top]], p[i])) --top;
		q[++top] = i; k[top] = slope(p[q[top - 1]], p[i]);
	}

	for(RG int i = top; i; --i)
		for(RG int j = p[q[i]].id; j; j = next[j]) ok[j] = 1;
	for(RG int i = 1; i <= n; i++) if(ok[i]) printf("%d ", i);
	return 0;
}