Hdu 2892 Area, hdu2892area

Hdu 2892 Area, hdu2892area

Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2892

Solution:

 

Returns the area of the intersection of a polygon and a circle.

 

Divides a Polygon into n triangles based on the center of the circle.

 

Next, we can find the area of the triangle and the circle.

 

 

 

Because one point of a triangle is the center of the circle, the other two points and circles of the triangle are as follows:

 

(1) The two points are in the circle, and the area of the triangle and the circle are equal to the area of the triangle.

 

(2) A point is located outside the circle, and a point is located in the circle. The area of the triangle and the circle intersection = Area of the small triangle + area of the slice

 

(3) The two points are both out of the circle and fall into several situations:

 

1. When a line segment composed of two points and the circle are 0 or 1 point, the area of the triangle and the circle is equal to the area of the slice.

 

2. When two points of the intersection of a line segment and a circle composed of two points, the area of the triangle and the Circle = large slice area + small triangle area-Small slice Area

 

1 # include <cmath> 2 # include <cstdio> 3 # include <vector> 4 # include <algorithm> 5 using namespace std; 6 7 # define MAXN 100000 + 10 8 # define PI acos (-1.0) 9 # define EPS 0.00000001 10 11 int dcmp (double x) {12 if (fabs (x) <EPS) 13 return 0; 14 return x <0? -1: 1; 15} 16 17 struct Point {18 double x, y; 19 Point (double x = 0, double y = 0): x (x ), y (y) {} 20}; 21 22 struct Circle {23 Point c; 24 double r; 25 Circle (Point c = Point (0, 0), double r = 0): c (c), r (r) {}26}; 27 28 typedef Point Vector; 29 30 Vector operator + (Vector A, Vector B) {31 return Vector (. x + B. x,. y + B. y); 32} 33 Vector operator-(Point A, Point B) {34 return Vector (. X-B. x,. y-B. y); 35} 36 Vector operator * (Vector A, double p) {37 return Vector (. x * p,. y * p); 38} 39 Vector operator/(Vector A, double p) {40 return Vector (. x/p,. y/p); 41} 42 43 double dot (Vector A, Vector B) {44 return. x * B. x +. y * B. y; 45} 46 47 double length (Vector A) {48 return sqrt (dot (A, A); 49} 50 51 double angle (Vector A, Vector B) {52 return acos (dot (A, B )/ Length (A)/length (B); 53} 54 55 double cross (Vector A, Vector B) {56 return. x * B. y-. y * B. x; 57} 58 59 Circle bomb; // coordinates and radius of bomb explosion 60 Point p [MAXN]; // Island Point 61 int n; // island Points 62 63 double point_line_distance (Point P, Point A, Point B) {// distance from A Point to A straight line 64 Vector AP = P-A, AB = B-; 65 return fabs (cross (AP, AB)/length (AB); 66} 67 68 Point point_line_projection (Point P, Point A, Point B) {// Point on A straight line Ing 69 Vector v = B-A; 70 return A + v * (dot (v, P-A)/dot (v, v )); 71} 72 73 int circle_line_intersect (Circle C, Point A, Point B, vector <Point> & v) {74 double dist = point_line_distance (C. c, A, B); 75 int d = dcmp (dist-C. r); 76 if (d> 0) {77 return 0; 78} 79 Point pro = point_line_projection (C. c, A, B); 80 if (d = 0) {81 v. push_back (pro); 82 return 1; 83} 84 double len = sqrt (C. r * C. r- Dist * dist); // ing theorem 85 Vector AB = B-A; 86 Vector l = AB/length (AB) * len; 87 v. push_back (pro + l); 88 v. push_back (pro-l); 89 return 2; 90} 91 92 bool point_on_segment (Point P, Point A, Point B) {// determine 93 Vector PA = A-P, PB = B-P; 94 return dcmp (cross (PA, PB )) = 0 & dcmp (dot (PA, PB) <= 0; 95} 96 97 double circle_delta_intersect_area (Circle C, Point A, Point B) {98 Vector CA = -C. c, CB = B-C. c; 99 double da = length (CA), db = length (CB); 100 101 da = dcmp (da-C. r), db = dcmp (db-C. r); 102 103 if (da <= 0 & db <= 0) {// 104 return fabs (cross (CA, CB) * 0.5 in the circle; 105} 106 107 vector <Point> v; 108 int num = circle_line_intersect (C, A, B, v); // The Relationship Between the circle and the straight line is 109 double carea = C. r * C. r * PI; 110 Point t; 111 if (da <= 0 & db> 0) {// point on the left in the circle the point on the right outside the circle 112 t = point_on_segment (v [0],, B )? V [0]: v [1]; 113 114 double area = fabs (cross (CA, t-C. c) * 0.5, an = angle (CB, t-C. c); 115 return area + carea * an/PI/2; 116} 117 if (da> 0 & db <= 0) {// The left point is outside the circle. the right point is in the circle. 118 t = point_on_segment (v [0], A, B )? V [0]: v [1]; 119 120 double area = fabs (cross (CB, t-C. c) * 0.5, an = angle (CA, t-C. c); 121 return area + carea * an/PI/2; 122} 123 // both vertices are out of the circle 124 if (num = 2) {125 double bigarea = carea * angle (CA, CB)/PI/2,126 smallarea = carea * angle (v [0]-C. c, v [1]-C. c)/PI/2,127 deltaarea = fabs (cross (v [0]-C. c, v [1]-C. c) * 0.5; 128 return bigarea + deltaarea-smallarea; 129} 130 return carea * Angle (CA, CB)/PI/2; // two vertices are both in the outer circle. One or two 131} 132 133 double circle_polygon_intersect_area () {// The Source polygon intersection area is 134 p [n] = p [0]; 135 double ans = 0; 136 for (int I = 0; I <n; I ++) {137 double area = circle_delta_intersect_area (bomb, p [I], p [I + 1]); 138 if (cross (p [I]-bomb. c, p [I + 1]-bomb. c) <0) {139 area =-area; 140} 141 ans + = area; 142} 143 return ans> 0? Ans:-ans; 144} 145 146 void solve () {147 scanf ("% d", & n); 148 for (int I = 0; I <n; I ++) {149 scanf ("% lf", & p [I]. x, & p [I]. y); 150} 151 printf ("%. 2lf \ n ", circle_polygon_intersect_area (); 152} 153 154 int main () {155 // freopen (" data. in "," r ", stdin); 156 double x, y, h, x1, y1, r; 157 while (~ Scanf ("% lf", & x, & y, & h) {158 scanf ("% lf", & x1, & y1, & r); 159 160 double t = sqrt (0.2 * h ); // h = 0.5 * G * t ^ 2 gravity Acceleration Formula 161 162 bomb = Circle (Point (x1 * t + x, y1 * t + y), r ); 163 164 solve (); 165} 166 return 0; 167}

 

Help me explain this question (HDU)

Bare BFS, but pay attention to the following mark, with x, y, t 3D mark. Because some coordinates have been used to reset the bomb multiple times ..
Paste my code ..
# Include <stdio. h>
Struct queue
{
Int x;
Int y;
Int time;
Int t;
} A [200000];
Int si, sj, ei, ej, n, m, d [4] [2] = {-, 0,-, 1 };
Int map [10] [10], hash [10] [10] [10];
Int bfs ()
{
Int I, x1, y1, t1, head, tail;
A [0]. x = si;
A [0]. y = sj;
A [0]. time = 0;
A [0]. t = 6;
Head = 0; tail = 1;
While (head <tail)
{
// Printf ("\ n % d \ n", a [head]. x, a [head]. y, a [head]. time, a [head]. t );
// Printf ("------------------------------------- \ n ");

For (I = 0; I <4; I ++)
{
If (a [head]. t <2)
Break;
X1 = a [head]. x + d [I] [0];
Y1 = a [head]. y + d [I] [1];
If (map [x1] [y1] = 4)
T1 = 6;
Else
T1 = a [head]. T-1;
If (hash [x1] [y1] [t1])
Continue;
If (x1 <0 | x1> = n | y1 <0 | y1> = m)
Continue;
If (map [x1] [y1] = 0)
Continue;
A [tail]. x = x1;
A [tail]. y = y1;
A [tail]. t = t1;
A [tail]. time = a [head]. time + 1;
Hash [x1] [y1] [a [tail]. t] = 1; // note the mark here.
// Printf ("% d \ n", a [tail]. x, a [tail]. y, a [tail]. time, a [tail]. t );
If (x1 = ei & y1 = ej)
Return (a [head]. time + 1 );
Tail ++;
}
// Printf ("\ n ");
Head ++;
}
Return 0;
} ...... Remaining full text>