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。

 

 

　　

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<set>
  7 #include<map>
  8 #include<stack>
  9 #include<time.h>
 10 #include<cstdlib>
 11 #include<cmath>
 12 #include<list>
 13 using namespace std;
 14 #define MAXN 5000006
 15 #define eps 1e-9
 16 #define For(i,a,b) for(int i=a;i<=b;i++)
 17 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 #define mkp make_pair
 21 #define pb push_back
 22 #define cr clear()
 23 #define sz size()
 24 #define met(a,b) memset(a,b,sizeof(a))
 25 #define iossy ios::sync_with_stdio(false)
 26 #define fr freopen
 27 #define pi acos(-1.0)
 28 #define Vector Point
 29 #define fir first
 30 #define sec second
 31 #define it_s_too_hard main
 32 #define I_can_t_solve_it solve
 33 //const long long inf=1LL<<62;
 34 const int inf=1e9+9;
 35 const int Mod=1e9+7;
 36 typedef unsigned long long ull;
 37 typedef long long ll;
 38 typedef pair<int,int> pii;
 39 typedef pair<ll,ll> pll;
 40 typedef double ld;
 41 inline int dcmp(ld x){ if(fabs(x)<=eps) return 0; return x<0?-1:1;}
 42 inline int scan(){
 43     int x=0,f=1;char ch=getchar();
 44     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 45     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 46     return x*f;
 47 }
 48 inline ll scan(ll x){
 49     int f=1;char ch=getchar();x=0;
 50     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 51     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 52     return x*f;
 53 }
 54 struct Point{
 55     int x,y;
 56     Point(int x=0,int y=0):x(x),y(y) {}
 57     Point operator - (const Point &a)const { return Point(x-a.x,y-a.y);}
 58     Point operator * (const ll &a)const{return Point(x*a,y*a);  }
 59     Point operator + (const Point &a)const{return Point(x+a.x,y+a.y);}
 60     bool operator == (const Point &a)const{ return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;}
 61     bool operator < (const Point &a)const{if(x==a.x) return y<a.y;return x<a.x;}
 62     void read(){scanf("%d%d",&x,&y);}
 63     void out(){cout<<x<<" "<<y<<endl;}
 64 };
 65 double Dot(Vector a,Vector b){
 66     return a.x*b.x+a.y*b.y;
 67 }
 68 double dis(Vector a){
 69     return sqrt(Dot(a,a));
 70 }
 71 ll Cross(Vector a,Vector b){
 72     return a.x*1LL*b.y-a.y*1LL*b.x;
 73 }
 74 bool chk(Point p1,Point p2,Point p3,Point p){
 75     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));
 76 }
 77 int n,m;
 78 Point p[5005],q[5005];
 79 int mp[505][505];
 80 void solve(){
 81     n=scan();m=scan();    
 82     met(mp,0);
 83     For(i,0,n-1) p[i].read();
 84     For(i,1,m) q[i].read();
 85     sort(p,p+n);
 86     For(i,0,n-1){
 87         For(j,i+1,n-1){
 88             if(p[i].x==p[j].x) continue;
 89             For(k,1,m){
 90                 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]++;
 91             }
 92         }
 93     }
 94     int ans=0;
 95     For(i,0,n-1){
 96         For(j,i+1,n-1){
 97             For(k,j+1,n-1){
 98                 int c1=mp[i][j],c2=mp[j][k],c3=mp[i][k];
 99                 if(c1+c2==c3) ans++;
100             }
101         }
102     }
103     cout<<ans<<endl;
104 }
105 int it_s_too_hard(){
106     int t=1;
107     For(i,1,t){
108         I_can_t_solve_it();
109     }
110     return 0;
111 }
蒻菜代码