CodeForces - 552D Vanya and Triangles （数学几何求三角形个数）水

CodeForces - 552D Vanya and Triangles Time Limit: 4000MS   Memory Limit: 524288KB   64bit IO Format: %I64d & %I64u Submit Status Description 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. Sample Input 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 Hint 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. Source Codeforces Round #308 (Div. 2) //题意： 给你n个点，问可以组成多少个三角形？ //思路： 先求出来所有的情况为sum=C(n,3)，再找出来不能组成的三角形的个数（三点在一条直线上）C(kk,3)，相减即为所求。 在求kk时会用到去重公式：kk=(sqrt(1+8*cnt)+1)/2;不明白的自己推一下就知道了。 #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<iostream> #define ll long long #define N 2010 using namespace std; struct zz { double x; double y; }p[N]; bool cmp1(zz a,zz b) { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } struct ss { double k; double b; }pp[N*N]; bool cmp2(ss a,ss b) { if(a.k==b.k) return a.b<b.b; return a.k<b.k; } ll C(int x,int k) { int i,j; ll s=1; for(i=x,j=1;i>x-k;i--,j++) s*=i,s/=j; return s; } int main() { int n,i,j,k; while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); sort(p,p+n,cmp1); k=0; double x,y; for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { x=(p[j].x-p[i].x);y=(p[j].y-p[i].y); if(x==0) { pp[k].k=1000.0; pp[k].b=p[j].x; } else { pp[k].k=y/x; pp[k].b=p[i].y-pp[k].k*p[i].x; } k++; } } sort(pp,pp+k,cmp2); int cnt=1,kk; ll num=C(n,3); for(i=0;i<k;i++) { if(fabs(pp[i].k-pp[i+1].k)<=1e-7&&fabs(pp[i].b-pp[i+1].b)<=1e-7) cnt++; else { // printf("%d\n",cnt); kk=(sqrt(1+8*cnt)+1)/2; num-=C(kk,3); cnt=1; } } printf("%lld\n",num); } return 0; }