BZOJ5317 JSOI2018部落战争（凸包）

　　即询问凸包是否有交。这显然可以直接求半平面交，但是复杂度O(q(n+m))，且没有什么优化空间。

　　更直接地表示，即相当于询问是否存在点a∈A，b∈B，使得a+d=b。移项，得到d=b-a。可以发现等式右边是一个闵可夫斯基和。求闵可夫斯基和只需要分别求出两个凸包，然后每次考虑a_i+1+b_i和a_i+b_i+1哪个将作为凸包中下一个点。将其求出后，只需要判断点是否在凸包内。二分找到上下边界即可。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
#define ll long long
#define vector point
#define N 200010
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,q;
const double eps=1E-;
struct point
{
    int x,y;
    vector operator +(const vector&a) const
    {
        return (vector){x+a.x,y+a.y};
    }
    vector operator -(const vector&a) const
    {
        return (vector){x-a.x,y-a.y};
    }
    ll operator *(const vector&a) const
    {
        return 1ll*x*a.y-1ll*y*a.x;
    }
    bool operator <(const point&a) const
    {
        return x<a.x||x==a.x&&y<a.y;
    }
}a[N],b[N],c[N],d[N],e[N],f[N];
struct line
{
    point a;vector p;
    double f(int x){return a.y+(double)(x-a.x)/p.x*p.y;}
};
void makehull(point *hull,point *a,int &n)
{
    sort(a+,a+n+);hull[]=a[];
    int m=;
    for (int i=;i<=n;i++)
    {
        while (m>&&(a[i]-hull[m-])*(hull[m]-hull[m-])>) m--;
        hull[++m]=a[i];
    }
    for (int i=n-;i>=;i--)
    {
        while (m>&&(a[i]-hull[m-])*(hull[m]-hull[m-])>) m--;
        hull[++m]=a[i];
    }
    n=m;
}
void merge(point *up,point *down,point *a,point *b,int &n,int &m)
{
    int p=,u=,v=;up[]=a[]+b[];
    while (u<n||v<m)
    {
        if (u==n) v++;
        else if (v==m) u++;
        else if ((a[u+]+b[v]-up[p])*(a[u]+b[v+]-up[p])>) u++;else v++;
        while (p>&&(a[u]+b[v]-up[p-])*(up[p]-up[p-])>) p--;
        up[++p]=a[u]+b[v];
    }
    for (int i=;i<=p;i++) if (up[i].x>up[i+].x) {n=i;break;}
    for (int i=n;i<=p;i++) down[i-n+]=up[i];m=p-n+;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5317.in","r",stdin);
    freopen("bzoj5317.out","w",stdout);
#endif
    n=read(),m=read(),q=read();
    for (int i=;i<=n;i++) a[i].x=read(),a[i].y=read();
    for (int i=;i<=m;i++) b[i].x=-read(),b[i].y=-read();
    makehull(c,a,n),makehull(d,b,m);
    merge(e,f,c,d,n,m);reverse(f+,f+m+);
    for (int i=;i<=q;i++)
    {
        int x=read(),y=read();
        int u=lower_bound(e+,e+n+,(point){x,y})-e;
        if (u==||u==n+||(line){e[u-],e[u]-e[u-]}.f(x)-eps>y) {printf("0\n");continue;}
        u=lower_bound(f+,f+m+,(point){x,y})-f;
        if (u==||u==m+||(line){f[u-],f[u]-f[u-]}.f(x)+eps<y) {printf("0\n");continue;}
        printf("1\n");
    }
    return ;
}