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LA 4064 Magnetic Train Tracks

时间:2016-08-04 06:49:37      阅读:260      评论:0      收藏:0      [点我收藏+]

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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:

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  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:

 

LA 4064 Magnetic Train Tracks

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原文地址:http://www.cnblogs.com/astoninfer/p/5735219.html

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