UVaLive 4064 Magnetic Train Tracks (极角排序)

题意：给定 n 个不三点共线的点，然后问你能组成多少锐角或者直角三角形。

析：可以反过来求，求有多少个钝角三角形，然后再用总的减去，直接求肯定会超时，但是可以枚举每个点，以该点为钝角的那个顶点，然后再枚举另一条边，维护与该边大于90度并小于等于180度的点的数量，这里要用极角排序，这样就可以减小时间复杂度，但是会WA，要控制精度，但是精度是个迷，表示不会。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1500 + 50;

const int maxm = 1e6 + 10;

const LL mod = 1000000000000000LL;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int a[maxn], b[maxn];

double p[maxn<<1];

int main(){

  int kase = 0;

  while(scanf("%d", &n) == 1 && n){

    for(int i = 0; i < n; ++i)  scanf("%d %d", a + i, b + i);

    int ans = 0;

    for(int i = 0; i < n; ++i){

      int cnt = 0;

      for(int j = 0; j < n; ++j)  if(i != j)

        p[cnt++] = atan2(b[j]-b[i], a[j]-a[i]);

      sort(p, p + cnt);

      for(int j = 0; j < cnt; ++j)  p[j+cnt] = p[j] + PI * 2.;

      int k = 0, l = 0;

      for(int j = 0; j < cnt; ++j){

        while(p[k] - p[j] + eps <= PI / 2.)  ++k;

        while(p[l] - p[j] + eps < PI)  ++l;

        ans += l - k;

      }

    }

    ans = n * (n-1) * (n-2) / 6 - ans;

    if(n < 3)  ans = 0;

    printf("Scenario %d:\n", ++kase);

    printf("There are %d sites for making valid tracks\n", ans);

  }

  return 0;

}