#308 (div.2) D. Vanya and Triangles

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1.题目描述：点击打开链接

2.解题思路：本题是一道简单的计算几何题，统计一个图中有多少个三角形，由于给的时间很宽，完全可以用O(N^3)的算法来解决，判断是否构成三角形只需要用向量来判断三点是否共线即可。

3.代码：

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef long long ll;typedef unsigned long long ull;#define me(s) memset(s,0,sizeof(s))#define For(i,n) for(int i=0;i<(n);i++)#define pb push_back#define sz size#define clr clear#define F(a,b) for(int i=a;b;i++)struct Point{    int x,y;    Point(int x=0,int y=0):x(x),y(y){}    Point operator +(const Point&p)const    {        return Point(x+p.x,y+p.y);    }    Point operator -(const Point&p)const    {        return Point(x-p.x,y-p.y);    }    void read()    {        scanf("%d%d",&x,&y);    }}a[2005];bool ok(int i,int j,int k){    Point e1=a[i]-a[j];    Point e2=a[i]-a[k];    if(e1.x*e2.y-e1.y*e2.x==0)return false;    return true;}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            a[i].read();        ll cnt=0;        for(int i=0;i<n;i++)            for(int j=i+1;j<n;j++)            for(int k=j+1;k<n;k++)            if(ok(i,j,k))            cnt++;        printf("%I64d\n",cnt);    }    return 0;}

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