[CodeForces]CodeForces 13D 几何 思维

　　大致题意：

　　　　给出N个红点和M个蓝点，问可以有多少个红点构成的三角形，其内部不含有蓝点

　　　假设我们现在枚举了一条线段（p[i],p[j]），我们可以记录线段下方满足(min（p[i].x,p[j].x）<x<=max（p[i].x,p[j].x） 的数量

　　　 时间复杂度为O（N*N*M） 

　　　那么我们就可以枚举三角形O(1)判断三角形内有无红点。

　　

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 5000006
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++)
#define Fore(i,a,b) for(int i=a;i>=b;i--)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fr freopen
#define pi acos(-1.0)
#define Vector Point
#define fir first
#define sec second
#define it_s_too_hard main
#define I_can_t_solve_it solve
//const long long inf=1LL<<62;
const int inf=1e9+9;
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef double ld;
inline int dcmp(ld x){ if(fabs(x)<=eps) return 0; return x<0?-1:1;}
inline int scan(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline ll scan(ll x){
    int f=1;char ch=getchar();x=0;
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct Point{
    int x,y;
    Point(int x=0,int y=0):x(x),y(y) {}
    Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
    Point operator * (const ll &a)const{return Point(x*a,y*a);  }
    Point operator + (const Point &a)const{return Point(x+a.x,y+a.y);}
    bool operator == (const Point &a)const{ return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;}
    bool operator < (const Point &a)const{if(x==a.x) return y<a.y;return x<a.x;}
    void read(){scanf("%d%d",&x,&y);}
    void out(){cout<<x<<" "<<y<<endl;}
};
double Dot(Vector a,Vector b){
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a){
    return sqrt(Dot(a,a));
}
ll Cross(Vector a,Vector b){
    return a.x*1LL*b.y-a.y*1LL*b.x;
}
bool chk(Point p1,Point p2,Point p3,Point p){
    return abs(Cross(p2-p,p1-p))+abs(Cross(p3-p,p2-p))+abs(Cross(p1-p,p3-p))==abs(Cross(p2-p1,p3-p1));
}
int n,m;
Point p[5005],q[5005];
int mp[505][505];
void solve(){
    n=scan();m=scan();    
    met(mp,0);
    For(i,0,n-1) p[i].read();
    For(i,1,m) q[i].read();
    sort(p,p+n);
    For(i,0,n-1){
        For(j,i+1,n-1){
            if(p[i].x==p[j].x) continue;
            For(k,1,m){
                if(q[k].x>p[i].x && q[k].x<=p[j].x && Cross(q[k]-p[j],p[i]-p[j])<0) mp[i][j]++;
            }
        }
    }
    int ans=0;
    For(i,0,n-1){
        For(j,i+1,n-1){
            For(k,j+1,n-1){
                int c1=mp[i][j],c2=mp[j][k],c3=mp[i][k];
                if(c1+c2==c3) ans++;
            }
        }
    }
    cout<<ans<<endl;
}
int it_s_too_hard(){
    int t=1;
    For(i,1,t){
        I_can_t_solve_it();
    }
    return 0;
}