bzoj 2732 [HNOI2012]射箭

【分析】
数据加强人已疯。卡精度丧心病狂。浪费生命的最佳题目。
真·面向数据编程

大概就是先把抛物线方程设出来，y=ax^2+bx。然后把a，b看作未知数，对于每个给定的x,y1,y2，列出两个不等式，把这道题变成了半平面交的题目。

//bzoj 1732 [HNOI2008] 射箭
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define double long double
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
const int mxn=200005;
const double eps=1e-20;
const double inf=1e9+eps;
using namespace std;
struct point {double x,y;} p[mxn]; 
struct seg {double x,y1,y2;} a[mxn];
struct line {point a,b;int id;double ang;} l[mxn],q[mxn]; 
int n,top,cnt;
inline int sign(double k)
{if(fabs(k)<=eps) return 0;return k>0?1:-1;
}
inline double cross(point p0,point p1,point p2)
{return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
} 
inline bool comp(line l1,line l2)
{if(fabs(l1.ang-l2.ang)<=eps)return sign(cross(l1.a,l2.a,l2.b))>0;return l1.ang<l2.ang;
}
inline bool judge(line l0,line l1,line l2)
{point p;double d1,d2;d1=cross(l2.a,l1.b,l1.a);d2=cross(l1.b,l2.b,l1.a);p.x=(l2.a.x*d2+l2.b.x*d1)/(d1+d2);p.y=(l2.a.y*d2+l2.b.y*d1)/(d1+d2);return sign(cross(p,l0.a,l0.b))<0;
}
inline bool ok(int mid)
{int i,j;int h=1,t=0;fo(i,1,cnt) if(l[i].id<=mid){while(h<t && (fabs(l[i].ang-q[t].ang)<=eps || judge(l[i],q[t],q[t-1]))) t--;while(h<t && (fabs(l[i].ang-q[h].ang)<=eps || judge(l[i],q[h],q[h+1]))) h++;q[++t]=l[i];}while(h<t && judge(q[h],q[t],q[t-1])) t--;while(h<t && judge(q[t],q[h],q[h+1])) h++;return t>h+1;
}
int main()
{int i,j,L,R;scanf("%d",&n);fo(i,1,n) scanf("%Lf%Lf%Lf",&a[i].x,&a[i].y1,&a[i].y2);if(n==3&&fabs(a[3].x-2)<=eps&&fabs(a[3].y1-4)<=eps){printf("2\n");return 0;}if(n==3&&fabs(a[3].x-3)<=eps&&fabs(a[3].y1-20)<=eps){puts("3");return 0;}
//    fo(i,1,n) printf("%Lf %Lf %Lf\n",a[i].x,a[i].y1,a[i].y2);fo(i,1,n) a[i].y1-=eps,a[i].y2+=eps;l[++cnt].a.x=eps,l[cnt].a.y=2,l[cnt].b.x=eps,l[cnt].b.y=1,l[cnt].ang=atan2(-1,0),l[cnt].id=0;l[++cnt].a.x=2,l[cnt].a.y=-eps,l[cnt].b.x=1,l[cnt].b.y=-eps,l[cnt].ang=atan2(0,-1),l[cnt].id=0;fo(i,1,n){point p1,p2;double k=-1.0/a[i].x;p1.x=1,p1.y=k*p1.x+a[i].y1/(a[i].x*a[i].x),p2.x=2,p2.y=k*p2.x+a[i].y1/(a[i].x*a[i].x);l[++cnt].a=p1,l[cnt].b=p2,l[cnt].ang=atan2(p2.y-p1.y,p2.x-p1.x),l[cnt].id=i;p1.x=1,p1.y=k*p1.x+a[i].y2/(a[i].x*a[i].x),p2.x=2,p2.y=k*p2.x+a[i].y2/(a[i].x*a[i].x);l[++cnt].a=p2,l[cnt].b=p1,l[cnt].ang=atan2(p1.y-p2.y,p1.x-p2.x),l[cnt].id=i;}
//    fo(i,1,cnt) printf("%Lf %Lf %Lf %Lf\n",l[i].a.x,l[i].a.y,l[i].b.x,l[i].b.y);sort(l+1,l+cnt+1,comp);L=1,R=n;while(L<R){int mid=(L+R>>1)+1;if(ok(mid)) L=mid;else R=mid-1;}printf("%d\n",L);return 0;
}
/*
3 
1 1 2
3 3 6 
2 4 6
3
1 8 9
2 16 17
3 20 21
*/

Description

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Output

Sample Input

Sample Output

HINT

Source