【BZOJ 1038】【ZJOI 2008】瞭望塔

http://www.lydsy.com/JudgeOnline/problem.php?id=1038

半平面交裸题,求完半平面后在折线段上的每个点竖直向上和半平面上的每个点竖直向下求距离,统计最小的值作为答案即可。

1A!!!斯巴达!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 303;

struct Point {
	double x, y;
	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
} t[N], p[N];
struct Line {
	Point p, v; double ang;
	Line(Point _p = Point(0, 0), Point _v = Point(0, 0), double _ang = 0) : p(_p), v(_v), ang(_ang) {}
	bool operator < (const Line &x) const {
		return ang < x.ang;
	}
} l[N], q[N];

Point operator + (Point a, Point b) {return Point(a.x + b.x, a.y + b.y);}
Point operator - (Point a, Point b) {return Point(a.x - b.x, a.y - b.y);}
Point operator * (Point a, double x) {return Point(a.x * x, a.y * x);}
Point operator / (Point a, double x) {return Point(a.x / x, a.y / x);}

int dcmp(double x) {return fabs(x) < 1e-8 ? 0 : (x < 0 ? -1 : 1);}
double Dot(Point a, Point b) {return a.x * b.x + a.y * b.y;}
double Cross(Point a, Point b) {return a.x * b.y - a.y * b.x;}
double sqr(double x) {return x * x;}
double dis(Point a, Point b) {return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}

bool onleft(Point a, Line b) {return dcmp(Cross(a - b.p, b.v)) < 0;}
Point intersection(Line a, Line b) {
	Point u; double t;
	u = a.p - b.p;
	t = Cross(b.v, u) / Cross(a.v, b.v);
	return a.p + (a.v * t);
}

int n, head = 1, tail = 2;
double ans;

void mkhalf() {
	q[1] = l[1]; q[2] = l[2];
	p[1] = intersection(q[1], q[2]);
	for(int i = 3; i < n; ++i) {
		while (head < tail && !onleft(p[tail - 1], l[i])) --tail;
		while (head < tail && !onleft(p[head], l[i])) ++head;
		q[++tail] = l[i];
		if (dcmp(Cross(q[tail].v, q[tail - 1].v) == 0)) {
			--tail;
			if (onleft(l[i].p, q[tail])) q[tail] = l[i];
		}
		if (head < tail) p[tail - 1] = intersection(q[tail - 1], q[tail]);
	}
//	while (head < tail + 1 && !onleft(p[tail - 1], q[head])) --tail;
}

double cal(int num) {
	Point j;
	double x = t[num].x, y = t[num].y;
	if (x <= p[head].x) {
		j = intersection(q[head], Line(t[num], Point(0, 1)));
		return j.y - y;
	}
	if (x >= p[tail - 1].x) {
		j = intersection(q[tail], Line(t[num], Point(0, 1)));
		return j.y - y;
	}
	int left = head, right = tail - 1, mid;
	while (left < right) {
		mid = (left + right + 1) >> 1;
		if (p[mid].x > x) right = mid - 1;
		else left = mid;
	}
	j = intersection(q[left + 1], Line(t[num], Point(0, 1)));
	return j.y - y;
}

double cal2(int num) {
	Point j;
	double x = p[num].x, y = p[num].y;
	int left = 1, right = n, mid;
	while (left < right) {
		mid = (left + right + 1) >> 1;
		if (t[mid].x > x) right = mid - 1;
		else left = mid;
	}
	j = intersection(Line(t[left], t[left + 1] - t[left]), Line(p[num], Point(0, -1)));
	return y - j.y;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%lf", &t[i].x);
	for(int i = 1; i <= n; ++i) scanf("%lf", &t[i].y);
	
	for(int i = 1; i < n; ++i) l[i] = Line(t[i], t[i + 1] - t[i]), l[i].ang = atan2(l[i].v.y, l[i].v.x);
	sort(l + 1, l + n);
	mkhalf();
	
	ans = cal(1);
	for(int i = 2; i <= n; ++i)
		ans = min(ans, cal(i));
	for(int i = head; i < tail; ++i)
		if (t[1].x <= p[i].x && p[i].x <= t[n].x)
			ans = min(ans, cal2(i));
	
//	for(int i = head; i < tail; ++i) printf("%.2lf %.2lf\n", p[i].x, p[i].y);
	
	printf("%.3lf\n", ans + 1e-8);
	return 0;
}

因为半平面不会围成一个"圈",所以我把最后"去除冗余"的部分注释掉了。最后输出答案加eps是听别人说的,不加会怎么样呢,我也不知道_(:з」∠)_

NOI 2017 Bless All
posted @ 2016-07-16 21:03  abclzr  阅读(230)  评论(0编辑  收藏  举报