If you see the inputs you'll see the points come in randomly . But from the figure it's obvious that we need to go from right to left (At least for the way I'm thinking) .Lets say right one is point p_1 and the left most one is point p_n .p_1 has highest x co-ordinate and p_n has the lowest.So,I'm gonna sort the points according to their x co-ordinates.Then,for p_1 and p_2 we must take the total distance (no line segments right to p_1-p_n).And then we will check which point has the greater y co-ordinate than p2 just left to it.Lets say that's point i.Then we find the intersection point of line segments p_i-p_(i-1) and the horizontal line passing through p2.We find the distance between p_i and intersection point and add that to answer and so on.
Code :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 | #include<bits/stdc++.h> using namespace std; #define INF 1e9 #define EPS 1e-9 #define PI acos(-1.0) // important constant; alternative #define PI (2.0 * acos(0.0)) double DEG_to_RAD(double d) { return d * PI / 180.0; } double RAD_to_DEG(double r) { return r * 180.0 / PI; } // struct point_i { int x, y; }; // basic raw form, minimalist mode struct point_i { int x, y; // whenever possible, work with point_i point_i() { x = y = 0; // default constructor *** Usage : point_i p(); } point_i(int _x, int _y) : x(_x), y(_y) {} }; // user-defined *** Usage : point_i p(5,6); struct point { double x, y; // only used if more precision is needed point() { x = y = 0.0; // default constructor } point(double _x, double _y) : x(_x), y(_y) {} // user-defined bool operator < (point other) const // override less than operator { if (fabs(x - other.x) > EPS) // useful for sorting return x > other.x; // first criteria , by x-coordinate // < max to min for this problem's purpose return y < other.y; //Not necessary in this problem } // second criteria, by y-coordinate // < min to max // use EPS (1e-9) when testing equality of two floating points bool operator == (point other) const { return (fabs(x - other.x) < EPS && (fabs(y - other.y) < EPS)); } }; // 2 points have same x and y co-ordinate double dist(point p1, point p2) // Euclidean distance { // hypot(dx, dy) returns sqrt(dx * dx + dy * dy) return hypot(p1.x - p2.x, p1.y - p2.y); } // return double // rotate p by theta degrees CCW w.r.t origin (0, 0) //Going CCW theta is equivalent to going ( CW 360 - theta ) degree point rotate(point p, double theta) //The input parameter theta will be provided in degrees //function returns a point { double rad = DEG_to_RAD(theta); // multiply theta with PI / 180.0 return point(p.x * cos(rad) - p.y * sin(rad), p.x * sin(rad) + p.y * cos(rad)); } struct line { double a, b, c; }; // a way to represent a line //as generally lines are constructed from 2 points . No such constructor as line(a,b,c) // the answer is stored in the third parameter (pass by reference) // //pass by reference syntax : func(int &x){...} {int x; func(x);} void pointsToLine(point p1, point p2, line &l) //Usage : line Ln ; pointsToLine(p1,p2,Ln); { if (fabs(p1.x - p2.x) < EPS) // vertical line is fine //same x co-ordinate { l.a = 1.0; l.b = 0.0; l.c = -p1.x; // default values } else { l.a = -(double)(p1.y - p2.y) / (p1.x - p2.x); l.b = 1.0; // IMPORTANT: we fix the value of b to 1.0 l.c = -(double)(l.a * p1.x) - p1.y; } } // not needed since we will use the more robust form: ax + by + c = 0 (see above) struct line2 { double m, c; }; // another way to represent a line int pointsToLine2(point p1, point p2, line2 &l) { if (abs(p1.x - p2.x) < EPS) // special case: vertical line { l.m = INF; // l contains m = INF and c = x_value l.c = p1.x; // to denote vertical line x = x_value return 0; // we need this return variable to differentiate result //for vertical line the function returns 0 } else { l.m = (double)(p1.y - p2.y) / (p1.x - p2.x); l.c = p1.y - l.m * p1.x; return 1; // l contains m and c of the line equation y = mx + c //returns 1 so non-vertical } } //Next few functions are just for line[a,b,c] representation bool areParallel(line l1, line l2) // check coefficients a & b { return (fabs(l1.a-l2.a) < EPS) && (fabs(l1.b-l2.b) < EPS); } bool areSame(line l1, line l2) // also check coefficient c { return areParallel(l1 ,l2) && (fabs(l1.c - l2.c) < EPS); } // returns true (+ intersection point) if two lines are intersect bool areIntersect(line l1, line l2, point &p) //pass by reference //saving intersection point to point p { if (areParallel(l1, l2)) return false; // no intersection // solve system of 2 linear algebraic equations with 2 unknowns p.x = (l2.b * l1.c - l1.b * l2.c) / (l2.a * l1.b - l1.a * l2.b); // special case: test for vertical line to avoid division by zero if (fabs(l1.b) > EPS) p.y = -(l1.a * p.x + l1.c); else p.y = -(l2.a * p.x + l2.c); return true; } struct vec { double x, y; // name: `vec' is different from STL vector vec(double _x, double _y) : x(_x), y(_y) {} }; vec toVec(point a, point b) // convert 2 points to vector a->b AB { return vec(b.x - a.x, b.y - a.y); } //same as thinking about a point w.r.t. origin vec scale(vec v, double s) // nonnegative s = [<1 .. 1 .. >1] { return vec(v.x * s, v.y * s); } // shorter__same__longer point translate(point p, vec v) // translate p according to v { return point(p.x + v.x , p.y + v.y); } // the point will move according to vector (vectors x,y component //would be added to point's x,y co-ordinate) // convert point and gradient/slope to line void pointSlopeToLine(point p, double m, line &l) //If the slope and a single point is known the line can be constructed { l.a = -m; // always -m l.b = 1; // always 1 l.c = -((l.a * p.x) + (l.b * p.y)); } // compute this //Closest point in l from p void closestPoint(line l, point p, point &ans) { line perpendicular; // perpendicular to l and pass through p if (fabs(l.b) < EPS) // special case 1: vertical line { ans.x = -(l.c); ans.y = p.y; return; } if (fabs(l.a) < EPS) // special case 2: horizontal line { ans.x = p.x; ans.y = -(l.c); return; } pointSlopeToLine(p, 1 / l.a, perpendicular); // normal line // intersect line l with this perpendicular line // the intersection point is the closest point areIntersect(l, perpendicular, ans); //areIntersect returns bool and stores intersection point in answer } // returns the reflection of point on a line . | . <-- reflection void reflectionPoint(line l, point p, point &ans) { point b; closestPoint(l, p, b); // similar to distToLine vec v = toVec(p, b); // create a vector ans = translate(translate(p, v), v); } // translate p twice double dot(vec a, vec b) { return (a.x * b.x + a.y * b.y); } double norm_sq(vec v) //returns the value^2 of the vector { return v.x * v.x + v.y * v.y; } // returns the distance from p to the line defined by // two points a and b (a and b must be different) // the closest point is stored in the 4th parameter (byref) double distToLine(point p, point a, point b, point &c) { // formula: c = a + u * ab vec ap = toVec(a, p), ab = toVec(a, b); double u = dot(ap, ab) / norm_sq(ab); c = translate(a, scale(ab, u)); // translate a to c return dist(p, c); } // Euclidean distance between p and c // returns the distance from p to the line segment ab defined by // two points a and b (still OK if a == b) // the closest point is stored in the 4th parameter (byref) double distToLineSegment(point p, point a, point b, point &c) { vec ap = toVec(a, p), ab = toVec(a, b); double u = dot(ap, ab) / norm_sq(ab); if (u < 0.0) { c = point(a.x, a.y); // closer to a return dist(p, a); } // Euclidean distance between p and a if (u > 1.0) { c = point(b.x, b.y); // closer to b return dist(p, b); } // Euclidean distance between p and b return distToLine(p, a, b, c); } // run distToLine as above double angle(point a, point o, point b) // returns angle aob in rad { vec oa = toVec(o, a), ob = toVec(o, b); return acos(dot(oa, ob) / sqrt(norm_sq(oa) * norm_sq(ob))); } double cross(vec a, vec b) { return a.x * b.y - a.y * b.x; } //// another variant //int area2(point p, point q, point r) { // returns 'twice' the area of this triangle A-B-c // return p.x * q.y - p.y * q.x + // q.x * r.y - q.y * r.x + // r.x * p.y - r.y * p.x; //} // note: to accept collinear points, we have to change the `> 0' // returns true if point r is on the left side of line pq bool ccw(point p, point q, point r) { return cross(toVec(p, q), toVec(p, r)) > 0; } // returns true if point r is on the same line as the line pq bool collinear(point p, point q, point r) { return fabs(cross(toVec(p, q), toVec(p, r))) < EPS; } vector<point>vp; int main() { int C;//C for cases cin>>C; while(C--) { int N;int num_of_p; cin>>N; num_of_p = N; while(N--) { double x,y; cin>>x>>y; point p_temp(x,y); vp.push_back(p_temp); } sort(vp.begin(),vp.end()); double total_dist = 0.00000000f; total_dist += dist(vp[0],vp[1]); int b; for(int a=1;a<num_of_p;){ for(b = a+1;b<num_of_p;b++) { point temp1,temp2,temp2_down; temp1 = vp[a];temp2 = vp[b]; temp2_down = vp[b-1]; if(temp2.y>temp1.y) { line l1,l2; point paral_2_x(0,temp1.y); pointsToLine(temp1,paral_2_x,l1); pointsToLine(temp2,temp2_down,l2); point P_intrsct; areIntersect(l1,l2,P_intrsct); total_dist += dist(temp2,P_intrsct); a = b; } } if(b==num_of_p) break; } printf("%.2lf\n",total_dist); vp.clear(); //This can be a killer , whenever using vector and there are multiple test cases clear vector } return 0; } |
কোন মন্তব্য নেই:
একটি মন্তব্য পোস্ট করুন