【25.33%】【codeforces 552D】Vanya and Triangles

time limit per test4 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.

Input
The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane.

Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output
In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Examples
input
4
0 0
1 1
2 0
2 2
output
3
input
3
0 0
1 1
2 0
output
1
input
1
1 1
output
0
Note
Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn’t form a single triangle.

【题目链接】:http://codeforces.com/contest/552/problem/D

【题解】

时限给的宽。
直接暴力枚举就可以了。
判断3条线是否相交
O(N^3)；
1500MS过。。
MAXN=2000;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 2e3+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

struct abc
{
    int x,y;
};

int n;
abc a[MAXN];

bool xj(int t1,int t2,int t3)
{
    int x1 = a[t1].x,y1 = a[t1].y;
    int x2 = a[t2].x,y2 = a[t2].y;
    int x3 = a[t3].x,y3 = a[t3].y;
    if ((x2-x1)*(y3-y2)-(y2-y1)*(x3-x2)==0) return true;
    else
        return false;
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i].x),rei(a[i].y);
    LL ans = 0;
    rep1(i,1,n-2)
        rep1(j,i+1,n-1)
            rep1(k,j+1,n)
                if (!xj(i,j,k))
                    ans++;
    printf("%I64d
",ans);
    return 0;
}