题意：给定平面上$n(3\leq n \leq 1200)$个无三点共线的点，问这些点组成了多少个锐角三角形。
分析：显然任意三点可构成三角形，而锐角三角形不如直角或钝角三角形容易计数，因为后者有且仅有一个角度大于等于$90^{\circ}$的特征角。
于是考虑固定平面上每一个顶点，也就是固定了钝角或直角三角形的一个特征顶点，将其余所有点按照极角排序，然后固定一条侧边，统计有多少条
边和该侧边夹角不小于$90^{\circ}$。这些边必然是连续的，可以使用区间统计的办法，用二分查找在$O(log(n))$时间内做到。
因此总的复杂度是$O(n^2log(n))$。
实际上如果用浮点数储存极角，然而用度数差判断会产生误差，下面的代码是采用分数来比较。
利用向量积的符号来决定侧边的分布范围，需要讨论某些值得符号。
代码：
1 #include <algorithm>
2 #include <cstdio>
3 #include <cstring>
4 #include <string>
5 #include <queue>
6 #include <map>
7 #include <set>
8 #include <ctime>
9 #include <cmath>
10 #include <iostream>
11 #include <assert.h>
12 #define PI acos(-1.)
13 #pragma comment(linker, "/STACK:102400000,102400000")
14 #define max(a, b) ((a) > (b) ? (a) : (b))
15 #define min(a, b) ((a) < (b) ? (a) : (b))
16 #define mp make_pair
17 #define st first
18 #define nd second
19 #define keyn (root->ch[1]->ch[0])
20 #define lson (u << 1)
21 #define rson (u << 1 | 1)
22 #define pii pair<int, int>
23 #define pll pair<ll, ll>
24 #define pb push_back
25 #define type(x) __typeof(x.begin())
26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
29 #define dbg(x) cout << x << endl
30 #define dbg2(x, y) cout << x << " " << y << endl
31 #define clr(x, i) memset(x, (i), sizeof(x))
32 #define maximize(x, y) x = max((x), (y))
33 #define minimize(x, y) x = min((x), (y))
34 using namespace std;
35 typedef long long ll;
36 const int int_inf = 0x3f3f3f3f;
37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
38 const int INT_INF = (int)((1ll << 31) - 1);
39 const double double_inf = 1e30;
40 const double eps = 1e-14;
41 typedef unsigned long long ul;
42 inline int readint(){
43 int x;
44 scanf("%d", &x);
45 return x;
46 }
47 inline int readstr(char *s){
48 scanf("%s", s);
49 return strlen(s);
50 }
51 //Here goes 2d geometry templates
52 struct Point{
53 double x, y;
54 Point(double x = 0, double y = 0) : x(x), y(y) {}
55 };
56 typedef Point Vector;
57 Vector operator + (Vector A, Vector B){
58 return Vector(A.x + B.x, A.y + B.y);
59 }
60 Vector operator - (Point A, Point B){
61 return Vector(A.x - B.x, A.y - B.y);
62 }
63 Vector operator * (Vector A, double p){
64 return Vector(A.x * p, A.y * p);
65 }
66 Vector operator / (Vector A, double p){
67 return Vector(A.x / p, A.y / p);
68 }
69 bool operator < (const Point& a, const Point& b){
70 return a.x < b.x || (a.x == b.x && a.y < b.y);
71 }
72 int dcmp(double x){
73 if(abs(x) < eps) return 0;
74 return x < 0 ? -1 : 1;
75 }
76 bool operator == (const Point& a, const Point& b){
77 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
78 }
79 double Dot(Vector A, Vector B){
80 return A.x * B.x + A.y * B.y;
81 }
82 double Len(Vector A){
83 return sqrt(Dot(A, A));
84 }
85 double Angle(Vector A, Vector B){
86 return acos(Dot(A, B) / Len(A) / Len(B));
87 }
88 double Cross(Vector A, Vector B){
89 return A.x * B.y - A.y * B.x;
90 }
91 double Area2(Point A, Point B, Point C){
92 return Cross(B - A, C - A);
93 }
94 Vector Rotate(Vector A, double rad){
95 //rotate counterclockwise
96 return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
97 }
98 Vector Normal(Vector A){
99 double L = Len(A);
100 return Vector(-A.y / L, A.x / L);
101 }
102 void Normallize(Vector &A){
103 double L = Len(A);
104 A.x /= L, A.y /= L;
105 }
106 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){
107 Vector u = P - Q;
108 double t = Cross(w, u) / Cross(v, w);
109 return P + v * t;
110 }
111 double DistanceToLine(Point P, Point A, Point B){
112 Vector v1 = B - A, v2 = P - A;
113 return abs(Cross(v1, v2)) / Len(v1);
114 }
115 double DistanceToSegment(Point P, Point A, Point B){
116 if(A == B) return Len(P - A);
117 Vector v1 = B - A, v2 = P - A, v3 = P - B;
118 if(dcmp(Dot(v1, v2)) < 0) return Len(v2);
119 else if(dcmp(Dot(v1, v3)) > 0) return Len(v3);
120 else return abs(Cross(v1, v2)) / Len(v1);
121 }
122 Point GetLineProjection(Point P, Point A, Point B){
123 Vector v = B - A;
124 return A + v * (Dot(v, P - A) / Dot(v, v));
125 }
126 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
127 //Line1:(a1, a2) Line2:(b1,b2)
128 double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
129 c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
130 return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
131 }
132 bool OnSegment(Point p, Point a1, Point a2){
133 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 -p)) < 0;
134 }
135 Vector GetBisector(Vector v, Vector w){
136 Normallize(v), Normallize(w);
137 return Vector((v.x + w.x) / 2, (v.y + w.y) / 2);
138 }
139
140 bool OnLine(Point p, Point a1, Point a2){
141 Vector v1 = p - a1, v2 = a2 - a1;
142 double tem = Cross(v1, v2);
143 return dcmp(tem) == 0;
144 }
145 struct Line{
146 Point p;
147 Vector v;
148 Point point(double t){
149 return Point(p.x + t * v.x, p.y + t * v.y);
150 }
151 Line(Point p, Vector v) : p(p), v(v) {}
152 };
153 struct Circle{
154 Point c;
155 double r;
156 Circle(Point c, double r) : c(c), r(r) {}
157 Circle(int x, int y, int _r){
158 c = Point(x, y);
159 r = _r;
160 }
161 Point point(double a){
162 return Point(c.x + cos(a) * r, c.y + sin(a) * r);
163 }
164 };
165 int GetLineCircleIntersection(Line L, Circle C, double &t1, double& t2, vector<Point>& sol){
166 double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
167 double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r;
168 double delta = f * f - 4 * e * g;
169 if(dcmp(delta) < 0) return 0;
170 if(dcmp(delta) == 0){
171 t1 = t2 = -f / (2 * e); sol.pb(L.point(t1));
172 return 1;
173 }
174 t1 = (-f - sqrt(delta)) / (2 * e); sol.pb(L.point(t1));
175 t2 = (-f + sqrt(delta)) / (2 * e); sol.pb(L.point(t2));
176 return 2;
177 }
178 double angle(Vector v){
179 return atan2(v.y, v.x);
180 //(-pi, pi]
181 }
182 int GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol){
183 double d = Len(C1.c - C2.c);
184 if(dcmp(d) == 0){
185 if(dcmp(C1.r - C2.r) == 0) return -1; //two circle duplicates
186 return 0; //two circles share identical center
187 }
188 if(dcmp(C1.r + C2.r - d) < 0) return 0; //too close
189 if(dcmp(abs(C1.r - C2.r) - d) > 0) return 0; //too far away
190 double a = angle(C2.c - C1.c); // angle of vector(C1, C2)
191 double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
192 Point p1 = C1.point(a - da), p2 = C1.point(a + da);
193 sol.pb(p1);
194 if(p1 == p2) return 1;
195 sol.pb(p2);
196 return 2;
197 }
198 int GetPointCircleTangents(Point p, Circle C, Vector* v){
199 Vector u = C.c - p;
200 double dist = Len(u);
201 if(dist < C.r) return 0;//p is inside the circle, no tangents
202 else if(dcmp(dist - C.r) == 0){
203 // p is on the circles, one tangent only
204 v[0] = Rotate(u, PI / 2);
205 return 1;
206 }else{
207 double ang = asin(C.r / dist);
208 v[0] = Rotate(u, -ang);
209 v[1] = Rotate(u, +ang);
210 return 2;
211 }
212 }
213 int GetCircleCircleTangents(Circle A, Circle B, Point* a, Point* b){
214 //a[i] store point of tangency on Circle A of tangent i
215 //b[i] store point of tangency on Circle B of tangent i
216 //six conditions is in consideration
217 int cnt = 0;
218 if(A.r < B.r) { swap(A, B); swap(a, b); }
219 int d2 = (A.c.x - B.c.x) * (A.c.x - B.c.x) + (A.c.y - B.c.y) * (A.c.y - B.c.y);
220 int rdiff = A.r - B.r;
221 int rsum = A.r + B.r;
222 if(d2 < rdiff * rdiff) return 0; // one circle is inside the other
223 double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x);
224 if(d2 == 0 && A.r == B.r) return -1; // two circle duplicates
225 if(d2 == rdiff * rdiff){ // internal tangency
226 a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++;
227 return 1;
228 }
229 double ang = acos((A.r - B.r) / sqrt(d2));
230 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang);
231 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang);
232 if(d2 == rsum * rsum){
233 //one internal tangent
234 a[cnt] = A.point(base);
235 b[cnt++] = B.point(base + PI);
236 }else if(d2 > rsum * rsum){
237 //two internal tangents
238 double ang = acos((A.r + B.r) / sqrt(d2));
239 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang + PI);
240 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang + PI);
241 }
242 return cnt;
243 }
244 Point ReadPoint(){
245 double x, y;
246 scanf("%lf%lf", &x, &y);
247 return Point(x, y);
248 }
249 Circle ReadCircle(){
250 double x, y, r;
251 scanf("%lf%lf%lf", &x, &y, &r);
252 return Circle(x, y, r);
253 }
254 //Here goes 3d geometry templates
255 struct Point3{
256 double x, y, z;
257 Point3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) {}
258 };
259 typedef Point3 Vector3;
260 Vector3 operator + (Vector3 A, Vector3 B){
261 return Vector3(A.x + B.x, A.y + B.y, A.z + B.z);
262 }
263 Vector3 operator - (Vector3 A, Vector3 B){
264 return Vector3(A.x - B.x, A.y - B.y, A.z - B.z);
265 }
266 Vector3 operator * (Vector3 A, double p){
267 return Vector3(A.x * p, A.y * p, A.z * p);
268 }
269 Vector3 operator / (Vector3 A, double p){
270 return Vector3(A.x / p, A.y / p, A.z / p);
271 }
272 double Dot3(Vector3 A, Vector3 B){
273 return A.x * B.x + A.y * B.y + A.z * B.z;
274 }
275 double Len3(Vector3 A){
276 return sqrt(Dot3(A, A));
277 }
278 double Angle3(Vector3 A, Vector3 B){
279 return acos(Dot3(A, B) / Len3(A) / Len3(B));
280 }
281 double DistanceToPlane(const Point3& p, const Point3 &p0, const Vector3& n){
282 return abs(Dot3(p - p0, n));
283 }
284 Point3 GetPlaneProjection(const Point3 &p, const Point3 &p0, const Vector3 &n){
285 return p - n * Dot3(p - p0, n);
286 }
287 Point3 GetLinePlaneIntersection(Point3 p1, Point3 p2, Point3 p0, Vector3 n){
288 Vector3 v = p2 - p1;
289 double t = (Dot3(n, p0 - p1) / Dot3(n, p2 - p1));
290 return p1 + v * t;//if t in range [0, 1], intersection on segment
291 }
292 Vector3 Cross(Vector3 A, Vector3 B){
293 return Vector3(A.y * B.z - A.z * B.y, A.z * B.x - A.x * B.z, A.x * B.y - A.y * B.x);
294 }
295 double Area3(Point3 A, Point3 B, Point3 C){
296 return Len3(Cross(B - A, C - A));
297 }
298 class cmpt{
299 public:
300 bool operator () (const int &x, const int &y) const{
301 return x > y;
302 }
303 };
304
305 int Rand(int x, int o){
306 //if o set, return [1, x], else return [0, x - 1]
307 if(!x) return 0;
308 int tem = (int)((double)rand() / RAND_MAX * x) % x;
309 return o ? tem + 1 : tem;
310 }
311 ////////////////////////////////////////////////////////////////////////////////////
312 ////////////////////////////////////////////////////////////////////////////////////
313 void data_gen(){
314 srand(time(0));
315 freopen("in.txt", "w", stdout);
316 int times = 100;
317 printf("%d\n", times);
318 while(times--){
319 int r = Rand(1000, 1), a = Rand(1000, 1), c = Rand(1000, 1);
320 int b = Rand(r, 1), d = Rand(r, 1);
321 int m = Rand(100, 1), n = Rand(m, 1);
322 printf("%d %d %d %d %d %d %d\n", n, m, a, b, c, d, r);
323 }
324 }
325
326 struct cmpx{
327 bool operator () (int x, int y) { return x > y; }
328 };
329 int debug = 1;
330 int dx[] = {0, 0, 1, 1};
331 int dy[] = {1, 0, 0, 1};
332 //-------------------------------------------------------------------------
333 const int maxn = 2e3 + 10;
334 pll a[maxn];
335 pll b1[maxn], b2[maxn];
336 pll b[maxn];
337 int k1, k2, k;
338 int n;
339 ll ans;
340 bool cmp(const pll &lhs, const pll &rhs){
341 return lhs.st * rhs.nd < rhs.st * lhs.nd;
342 }
343 void solve(){
344 ans = 0;
345 FOR(i, 1, n){
346 ll yb_leq_0 = 0, yb_geq_0 = 0, yb_eqt_0 = 0;
347 ll yb_eqt_0_xb_geq_0 = 0, yb_eqt_0_xb_leq_0 = 0;
348 k1 = k2 = k = 0;
349 FOR(j, 1, n){
350 if(j == i) continue;
351 ll lhs = a[j].st - a[i].st, rhs = a[j].nd - a[i].nd;
352 yb_leq_0 += rhs <= 0, yb_geq_0 += rhs >= 0, yb_eqt_0 += rhs == 0;
353 if(rhs < 0) b1[k1++] = mp(lhs, rhs);
354 else if(rhs > 0) b2[k2++] = mp(lhs, rhs);
355 else if(lhs >= 0) ++yb_eqt_0_xb_geq_0;
356 else if(lhs <= 0) ++yb_eqt_0_xb_leq_0;
357 b[k++] = mp(lhs, rhs);
358 }
359 FOR(j, 0, k1 - 1) b1[j].st = -b1[j].st, b1[j].nd = -b1[j].nd;
360 sort(b1, b1 + k1, cmp), sort(b2, b2 + k2, cmp);
361 ll cnt = 0;
362 FOR(j, 0, k - 1){
363 ll xa = b[j].st, ya = b[j].nd;
364 if(!xa){
365 if(ya < 0) cnt += yb_geq_0;
366 else if(ya > 0) cnt += yb_leq_0;
367 else if(!ya) cnt += yb_geq_0 + yb_leq_0 + yb_eqt_0;
368 continue;
369 }else{
370 if(xa < 0) cnt += yb_eqt_0_xb_geq_0;
371 else if(xa > 0) cnt += yb_eqt_0_xb_leq_0;
372 pll tem = mp(-ya, xa);
373 if(xa < 0) tem.st = -tem.st, tem.nd = -tem.nd;
374 if(xa < 0){
375 int l = lower_bound(b2, b2 + k2, tem, cmp) - b2;
376 cnt += k2 - l;
377 int r = upper_bound(b1, b1 + k1, tem, cmp) - b1;
378 cnt += r;
379 }else if(xa > 0){
380 int l = lower_bound(b1, b1 + k1, tem, cmp) - b1;
381 cnt += k1 - l;
382 int r = upper_bound(b2, b2 + k2, tem, cmp) - b2;
383 cnt += r;
384 }
385 }
386 }
387 ans += cnt >> 1;
388 //printf("at %lld %lld count %lld\n", a[i].st, a[i].nd, cnt);
389 }
390 ans = (ll)n * (n - 1) * (n - 2) / 6 - ans;
391 }
392
393 //-------------------------------------------------------------------------
394 int main(){
395 //data_gen(); return 0;
396 //C(); return 0;
397 debug = 0;
398 ///////////////////////////////////////////////////////////////////////////////////////////////////////////////
399 if(debug) freopen("in.txt", "r", stdin);
400 //freopen("in.txt", "w", stdout);
401 int kase = 0;
402 while(~scanf("%d", &n) && n){
403 FOR(i, 1, n) scanf("%lld%lld", &a[i].st, &a[i].nd);
404 solve();
405 printf("Scenario %d:\nThere are %lld sites for making valid tracks\n", ++kase, ans);
406 }
407 //////////////////////////////////////////////////////////////////////////////////////////////////////////////
408 return 0;
409 }
code：
时间： 2024-10-11 07:47:34