uva109求凸包面积,判断点是不是在凸包内

自己想了一个方法判断点是不是在凸包内,先求出凸包面积,在求由点与凸包上每两个点之间的面积(点已经排好序了),如果两者相等,则点在凸包内,否则不在(时间复杂度可能有点高)但是这题能过

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-7;
const int N=100+10,maxn=500+100,inf=0x3f3f3f;

struct point{
    double x,y;
};
point p[N],s[N][N];
int n,top[N];
bool vis[N];
inline bool zero(double x)
{
    return fabs(x)<eps;
}
double dis(point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double dir(point p1,point p2,point p3)
{
    return (p3.x-p1.x)*(p2.y-p1.y)-(p3.y-p1.y)*(p2.x-p1.x);
}
bool comp(point p1,point p2)
{
    double te=dir(p[0],p1,p2);
    if(te<0)return 1;
    if(zero(te)&&dis(p[0],p1)<dis(p[0],p2))return 1;
    return 0;
}
void Graham(int k)
{
    int pos;
    double minx=inf,miny=inf;
    for(int i=0;i<n;i++)
    {
        if(p[i].x<minx||(p[i].x==minx&&p[i].y<miny))
        {
            minx=p[i].x;
            miny=p[i].y;
            pos=i;
        }
    }
    swap(p[0],p[pos]);
    sort(p+1,p+n,comp);
    p[n]=p[0];
    s[k][0]=p[0],s[k][1]=p[1], s[k][2]=p[2];
    top[k]=2;
    for(int i=3;i<=n;i++)
    {
        while(top[k]>=2&&dir(s[k][top[k]-1],s[k][top[k]],p[i])>=0)top[k]--;
        s[k][++top[k]]=p[i];
    }
}
double square(int k)
{
    double ans=0;
    for(int i=0;i<top[k]-1;i++)
        ans-=dir(s[k][0],s[k][i],s[k][i+1]);
    return ans/2;
}
bool inside(double sq,double x,double y,int k)
{
     double ans=0;
     point p0;p0.x=x,p0.y=y;
     for(int i=0;i<top[k];i++)
     {
         if(dir(p0,s[k][i],s[k][i+1])<0)ans-=dir(p0,s[k][i],s[k][i+1]);
         else ans+=dir(p0,s[k][i],s[k][i+1]);
     }
     return sq*2==ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int cnt=0;
    while(cin>>n,n!=-1){
        for(int i=0;i<n;i++)
            cin>>p[i].x>>p[i].y;
        Graham(cnt);
        cnt++;
    }
  /*  for(int i=0;i<cnt;i++)
    {
        for(int j=0;j<top[i];j++)
            cout<<s[i][j].x<<" "<<s[i][j].y<<endl;
    }*/
    memset(vis,0,sizeof vis);
    int x,y;
    double ans=0;
    while(cin>>x>>y){
        for(int i=0;i<cnt;i++)
        {
            double sq=square(i);
            if(!vis[i]&&inside(sq,x,y,i))
            {
                vis[i]=1;
                ans+=sq;
                break;
            }
        }
    }
    cout<<setiosflags(ios::fixed)<<setprecision(2)<<ans<<endl;
    return 0;
}
/*********************

*********************/
View Code
原文地址:https://www.cnblogs.com/acjiumeng/p/7197443.html