UVA 13024: Saint John Festival（凸包+二分 ，判定多个点在凸包内）

题意：给定N个点，Q次询问，问当前点知否在N个点组成的凸包内。

思路：由于是凸包，我们可以利用二分求解。

二分思路1：求得上凸包和下凸包，那么两次二分，如果点在对应上凸包的下面，对应下凸包的上面，那么在凸包内。

二分思路2：求得凸包(N)，划分为N-2个三角形，二分求得对应位置，验证是否在三角形内。

（如果不是凸包，则不能这样做。

三角形代码：

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=100010;
struct point{
    ll x,y;
    point(){}
    point(ll xx,ll yy):x(xx),y(yy){}
};
ll dot(point w,point v){ return w.x*v.x+w.y*v.y;}
ll det(point w,point v){ return w.x*v.y-w.y*v.x;}
point operator -(point w,point v){ return point(w.x-v.x,w.y-v.y);}
point a[maxn],ch[maxn]; int ttop,top,N,Q,ans;
bool cmp(point w,point v){
    if(w.x!=v.x) return w.x<v.x; return w.y<v.y;
}void convex()
{
    top=0;
    sort(a+1,a+N+1,cmp);
    rep(i,1,N) {
        while(top>1&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
    ttop=top;
    for(int i=N-1;i>=1;i--){
        while(top>ttop&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
}
bool check(point A)
{
    int L=2,R=top-2,Mid;
    while(L<=R){
        Mid=(L+R)>>1;
        if(det(ch[Mid]-ch[1],A-ch[1])<0) R=Mid-1;
        else {
            if(det(ch[Mid+1]-ch[1],A-ch[1])<=0&&det(ch[Mid+1]-ch[Mid],A-ch[Mid])>=0)
                return true;
            L=Mid+1;
        }
    }
    return false;
}
int main()
{
    while(~scanf("%d",&N)&&N){
        rep(i,1,N) scanf("%lld %lld",&a[i].x,&a[i].y);
        convex();  ans=0;
        scanf("%d",&Q);
        rep(i,1,Q) {
            point fcy;
            scanf("%lld %lld",&fcy.x,&fcy.y);
            if(check(fcy)) ans++;
        }
        printf("%d
",ans);
    }
    return 0;
}

上下凸包代码：不知道咋的，一直wa1，也有可能是思路有问题吧，日后再补。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=100010;
struct point{
    ll x,y;
    point(){}
    point(ll xx,ll yy):x(xx),y(yy){}
};
ll dot(point w,point v){ return w.x*v.x+w.y*v.y;}
ll det(point w,point v){ return w.x*v.y-w.y*v.x;}
point operator -(point w,point v){ return point(w.x-v.x,w.y-v.y);}
point a[maxn],ch[maxn]; int ttop,top,N,Q,ans;
bool cmp(point w,point v){
    if(w.x!=v.x) return w.x<v.x; return w.y<v.y;
}
void convex()
{
    top=0;
    sort(a+1,a+N+1,cmp);
    rep(i,1,N) {
        while(top>1&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
    ttop=top;
    for(int i=N-1;i>=1;i--){
        while(top>ttop&&det(ch[top]-ch[top-1],a[i]-ch[top-1])<=0) top--;
        ch[++top]=a[i];
    }
}
bool check1(point A)
{
    int L=1,R=ttop-1,Mid;
    while(L<=R){
        Mid=(L+R)>>1;
        if(A.x<ch[Mid].x) R=Mid-1;
        else {
            if(ch[Mid+1].x>=A.x&&det(ch[Mid+1]-ch[Mid],A-ch[Mid])>=0){
                    return true;
            }
            L=Mid+1;
        }
    }
    return false;
}
bool check2(point A)
{
    int L=ttop,R=top-1,Mid;
    while(L<=R){
        Mid=(L+R)>>1;
        if(A.x>ch[Mid].x) R=Mid-1;
        else {
            if(ch[Mid+1].x<=A.x&&det(ch[Mid+1]-ch[Mid],A-ch[Mid])>=0) return true;
            L=Mid+1;
        }
    }
    return false;
}
int main()
{
    while(~scanf("%d",&N)&&N){
        rep(i,1,N) scanf("%lld %lld",&a[i].x,&a[i].y);
        convex();  ans=0;
        scanf("%d",&Q);
        rep(i,1,Q) {
            point fcy;
            scanf("%lld %lld",&fcy.x,&fcy.y);
            if(check1(fcy)&&check2(fcy)) ans++;
        }
        printf("%d
",ans);
    }
    return 0;
}