BZOJ-2732 [HNOI2012]射箭

依旧转成不等式组,然后半平面交。

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <algorithm>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define maxn 234567
#define double long double
#define linf 1e15
using namespace std;
typedef long long ll;
inline int read()
{
	int x=0, f=1; char ch=getchar();
	while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while (isdigit(ch)) x=x*10+ch-'0', ch=getchar();
	return x*f;
}
struct P{double x, y;};
struct line{P a, b; int id; double ang;} l[maxn], a[maxn], q[maxn];
P operator - (P a, P b){return (P){a.x-b.x, a.y-b.y};}
double operator * (P a, P b){return a.x*b.y-a.y*b.x;}
bool operator < (line a, line b){if (a.ang==b.ang) return (a.b-a.a)*(b.b-b.a)>0; return a.ang<b.ang;}



int n, m, L, R, cnt;
inline double cal(double a, double b, int x){return b/a-a*x;}
P inter(line a, line b)
{
	double k1=(b.b-a.a)*(a.b-a.a), k2=(a.b-a.a)*(b.a-a.a), t=k2/(k1+k2);
	return (P){b.a.x+t*(b.b.x-b.a.x), b.a.y+t*(b.b.y-b.a.y)};
}
bool jud(line a, line b, line t){return (inter(a, b)-t.a)*(t.b-t.a)>0;}
void hpi(int x)
{
	int cnt=0;
	rep(i, 1, m) if (l[i].id<=x)
	{
		if (l[i].ang!=a[cnt].ang) cnt++;
		a[cnt]=l[i];
	}
	L=1, R=0;
	q[++R]=a[1]; q[++R]=a[2];
	rep(i, 3, cnt)
	{
		while (L<R && jud(q[R-1], q[R], a[i])) R--;
		while (L<R && jud(q[L+1], q[L], a[i])) L++;
		q[++R]=a[i];
	}
	while (L<R && jud(q[R-1], q[R], q[L])) R--;
	while (L<R && jud(q[L+1], q[L], q[R])) L++;
}


int main()
{
	n=read();
	l[++m].a=(P){-linf, -linf}; l[m].b=(P){linf, -linf};
	l[++m].a=(P){linf, -linf}; l[m].b=(P){linf, linf};
	l[++m].a=(P){linf, linf}; l[m].b=(P){-linf, linf};
	l[++m].a=(P){-linf, linf}; l[m].b=(P){-linf, -linf};
	rep(i, 1, n)
	{
		double x=read(), ya=read(), yb=read();
		l[++m].a.x=-1; l[m].a.y=cal(x, ya, -1); l[m].b.x=1; l[m].b.y=cal(x, ya, 1);
		l[++m].a.x=1; l[m].a.y=cal(x, yb, 1); l[m].b.x=-1; l[m].b.y=cal(x, yb, -1);
		l[m].id=l[m-1].id=i;
	}
	rep(i, 1, m) l[i].ang=atan2(l[i].b.y-l[i].a.y, l[i].b.x-l[i].a.x);
	sort(l+1, l+m+1);
	int l=0, r=n;
	while (l<r)
	{
		int mid=(l+r)>>1;
		hpi(mid+1);
		if (R-L>1) l=mid+1; else r=mid;
	}
	printf("%d\n", l);
	return 0;
}
posted @ 2015-05-05 18:45  NanoApe  阅读(183)  评论(0编辑  收藏  举报
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