UVA1475 Jungle Outpost [半平面交]

这个题就是一个半平面交加二分，因为敌人每次如果炸k个瞭望塔，那么连续的k个肯定比分散的k个要更优，所以每次二分答案k，将点每隔k个连一条新的线，判断这些新的线有无半平面交即可。

下面上代码

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define db double
using namespace std;
const int M=1e5+10;
const db eps=1e-7;
struct point{
    db x,y;
    point(db a=0,db b=0):x(a),y(b){}
    void in(){scanf("%lf%lf",&x,&y);}
}pp[M];

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,db    b){return point(a.x*b  ,a.y*b  );}
point operator /(point a,db    b){return point(a.x/b  ,a.y/b  );}

db cross(point a,point b){return a.x*b.y-a.y*b.x;}
db dot  (point a,point b){return a.x*b.x+a.y*b.y;}

struct line{
    point p,v;
    db ang;
    line(){}
    line(point a,point b):p(a),v(b){ang=atan2(v.y,v.x);}
    bool operator <(line a)const{return ang<a.ang;}
    point getp(db t){return p+v*t;}
}l[M];

bool onleft(line le,point p){
    return cross(le.v,p-le.p)>0;
}

point getlinecut(line a,line b){
    point u=a.p-b.p;
    db t=cross(b.v,u)/cross(a.v,b.v);
    return a.getp(t);
}

int fi,la;
point p[M];
line  q[M];
int n;
void init(int mid){for(int i=0;i<n;i++){l[i]=line(pp[i],pp[i]-pp[(i+mid+1)%n]);}}
//二分的mid，每隔mid个连一条线
bool halfcut(int mid){
    init(mid);
    sort(l,l+n);
    q[fi=la=0]=l[0];
    for(int i=1;i<n;i++){
        while(fi<la&&!onleft(l[i],p[la-1])) --la;
        while(fi<la&&!onleft(l[i],p[fi  ])) ++fi;
        q[++la]=l[i];
        if(fabs(cross(q[la].v,q[la-1].v))<eps){
            --la;if(onleft(q[la],l[i].p)) q[la]=l[i];
        }
        if(fi<la) p[la-1]=getlinecut(q[la],q[la-1]);
    }
    while(fi<la&&!onleft(q[fi],p[la-1])) --la;
    if(la-fi<=1) return 0;return 1;
}

int le,ri;
int main()
{
    while(scanf("%d",&n)==1&&n){
        for(int i=0;i<n;i++) pp[i].in();
        le=1,ri=n;
        while(le<=ri){
            int mid=(le+ri)>>1;
            if(halfcut(mid)) le=mid+1;
            else ri=mid-1;
        }
        printf("%d\n",le);
    }
    return 0;
}