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;
}

原文地址：https://www.cnblogs.com/dwtfukgv/p/8970060.html