FZU 2110 Star 数学

Star

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

FZU 2110

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


套一个锐角三角形的性质就可以了 
两边平方和大于第三边的平方和

开始用第一种方法来做 比较耗时 后来看了高端玩家是怎么循环的 第二种方法的耗时大大的降低
way1

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;

const int MX = 100 + 5;

struct Point{
    double x, y;
}P[MX];


bool check(int i, int j, int k){
    if(i == j || i == k || j == k)  return false;
    double len1 = (P[i].x - P[j].x)*(P[i].x - P[j].x) + (P[i].y - P[j].y)*(P[i].y - P[j].y);
    double len2 = (P[i].x - P[k].x)*(P[i].x - P[k].x) + (P[i].y - P[k].y)*(P[i].y - P[k].y);
    double len3 = (P[j].x - P[k].x)*(P[j].x - P[k].x) + (P[j].y - P[k].y)*(P[j].y - P[k].y);
    len1 = sqrt(len1);
    len2 = sqrt(len2);
    len3 = sqrt(len3);
    if(len1 * len1 + len2 * len2 > len3 * len3)
        if(len1 * len1 + len3 * len3 > len2 * len2)
             if(len3 * len3 + len2 * len2 > len1 * len1)
                  return true;

    return false;
}
int main(){
    //FIN
    int t;
    while (~scanf ("%d", &t)){
        while (t--){
            int n;
            scanf ("%d", &n);
            for (int i = 0; i < n; i ++) scanf ("%lf%lf", &P[i].x, &P[i].y);
            int cnt = 0;
            for (int i = 0; i < n; i ++){
                for(int j = 0; j < n ; j ++){
                    for(int k = 0; k < n; k ++){
                        if(check(i, j, k))  cnt ++;
                    }
                }
            }
            printf ("%d
",cnt/6);
        }
    }
    return 0;
}

way2

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;

const int MX = 100 + 5;

struct Point{
    double x, y;
}P[MX];


bool check(int i, int j, int k){
    double len1 = (P[i].x - P[j].x)*(P[i].x - P[j].x) + (P[i].y - P[j].y)*(P[i].y - P[j].y);
    double len2 = (P[i].x - P[k].x)*(P[i].x - P[k].x) + (P[i].y - P[k].y)*(P[i].y - P[k].y);
    double len3 = (P[j].x - P[k].x)*(P[j].x - P[k].x) + (P[j].y - P[k].y)*(P[j].y - P[k].y);
    len1 = sqrt(len1);
    len2 = sqrt(len2);
    len3 = sqrt(len3);
    if(len1 * len1 + len2 * len2 > len3 * len3)
        if(len1 * len1 + len3 * len3 > len2 * len2)
             if(len3 * len3 + len2 * len2 > len1 * len1)
                  return true;

    return false;
}
int main(){
    //FIN
    int t;
    while (~scanf ("%d", &t)){
        while (t--){
            int n;
            scanf ("%d", &n);
            for (int i = 0; i < n; i ++) scanf ("%lf%lf", &P[i].x, &P[i].y);
            int cnt = 0;
            for (int i = 0; i < n - 2; i ++){
                for(int j = i + 1; j < n - 1; j ++){
                    for(int k = j + 1; k < n; k ++){
                        if(check(i, j, k))  cnt ++;
                    }
                }
            }
            printf ("%d
",cnt);
        }
    }
    return 0;
}