ACM: FZU 2110 Star - 数学几何 - 水题

FZU 2110  Star

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Practice

Description

Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.

Input

The first line of the input contains an integer T (T≤10), indicating the number of test cases.

For each test case:

The first line contains one integer n (1≤n≤100), the number of stars.

The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct.

Output

For each test case, output an integer indicating the total number of different acute triangles.

Sample Input

1
3
0 0
10 0
5 1000

Sample Output

1

求锐角三角形的个数；

数学结论 ： 锐角三角形 任意边满足 a^2<b^2+c^2 ;

AC代码：

#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"cstdio"
#include"string"
#include"vector"
#include"stack"
#include"queue"
#include"cmath"
#include"map"
using namespace std;
typedef long long LL ;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FK(x) cout<<"["<<x<<"]\n"
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define bigfor(T)  for(int qq=1;qq<= T ;qq++)

struct Star {
	LL x;
	LL y;
} st[200];

bool cmp(LL a,LL b) {
	return a>b;
}

bool check(int i,int j,int k) {
	LL s[3];
	s[0]=(st[i].x-st[j].x)*(st[i].x-st[j].x)+(st[i].y-st[j].y)*(st[i].y-st[j].y);
	s[1]=(st[i].x-st[k].x)*(st[i].x-st[k].x)+(st[i].y-st[k].y)*(st[i].y-st[k].y);
	s[2]=(st[k].x-st[j].x)*(st[k].x-st[j].x)+(st[k].y-st[j].y)*(st[k].y-st[j].y);
//	FK(s[0]);
//	FK(s[1]);
//	FK(s[2]);
	sort(s,s+3,cmp);
	if(s[0]<s[1]+s[2])return 1;
	return 0;
}

int main() {
	int T;
	int n;
	scanf("%d",&T);
	bigfor(T) {
		scanf("%d",&n);
		for(int i=0; i<n; i++) {
			scanf("%I64d%I64d",&st[i].x,&st[i].y);
		}
		LL ans=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(check(i,j,k)) ans++;
				}
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
}