JOJ 2785 赛车 （半平面交）
Jilin University Online Judge System--2785:赛车
吉林大学OJ上的题目。中文题。
这是经典的半平面交的模型。直接套用半平面交的模板，不过要注意，对有向直线排序的时候要用叉积来比较，不然精度会丢失，从而导致排序出错。
代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <cmath> 6 7 using namespace std; 8 9 const double PI = acos(-1.0); 10 const double EPS = 1e-10; 11 inline int sgn(double x) { return (EPS < x) - (x < -EPS);} 12 struct Point { 13 double x, y; 14 Point() {} 15 Point(double x, double y) : x(x), y(y) {} 16 bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && sgn(y - a.y) < 0;} 17 bool operator == (Point a) const { return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;} 18 } ; 19 typedef Point Vec; 20 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);} 21 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);} 22 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);} 23 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);} 24 inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;} 25 inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;} 26 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);} 27 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));} 28 inline double ptDis(Point a, Point b) { return vecLen(a - b);} 29 inline Vec normal(Vec x) { 30 double len = vecLen(x); 31 return Vec(-x.y / len, x.x / len); 32 } 33 inline Vec vecUnit(Vec x) { return x / vecLen(x);} 34 35 struct DLine { 36 Point p; 37 Vec v; 38 int id; 39 double ang; 40 DLine() {} 41 DLine(Point p, Vec v) : p(p), v(v) { ang = atan2(v.y, v.x);} 42 bool operator < (const DLine &L) const { return sgn(ang - L.ang) < 0 || sgn(ang - L.ang) == 0 && p < L.p;} 43 bool operator == (const DLine &L) const { return p == L.p && v == L.v;} 44 } ; 45 46 inline bool onLeft(DLine L, Point p) { return sgn(crossDet(L.v, p - L.p)) >= 0;} 47 Point dLineIntersect(DLine a, DLine b) { 48 Vec u = a.p - b.p; 49 double t = crossDet(b.v, u) / crossDet(a.v, b.v); 50 return a.p + a.v * t; 51 } 52 53 const int N = 11111; 54 Point p[N]; 55 DLine q[N], rec[N]; 56 int ans[N], ansEnd; 57 58 bool cmp(DLine a, DLine b) { 59 return crossDet(a.v, b.v) > 0.0; 60 } 61 62 void halfPlane(DLine *L, int n) { 63 ansEnd = 0; 64 sort(L, L + n, cmp); 65 // for (int i = 0; i < n; i++) cout << L[i].id << ' '; cout << endl; 66 int fi, la; 67 q[fi = la = 0] = L[0]; 68 for (int i = 1; i < n; i++) { 69 // cout << fi << "=1=" << la << endl; 70 // cout << L[i].v.x << ' ' << L[i].v.y << endl; 71 // cout << (L[i].p - p[la - 1]).x << ' ' << (L[i].p - p[la - 1]).y << endl; 72 while (fi < la && !onLeft(L[i], p[la - 1]) && sgn(crossDet(L[i].v, L[i].p - p[la - 1]))) la--; 73 while (fi < la && !onLeft(L[i], p[fi]) && sgn(crossDet(L[i].p, L[i].p + L[i].v, p[fi]))) fi++; 74 // cout << fi << "=2=" << la << endl; 75 q[++la] = L[i]; 76 if (fabs(crossDet(q[la].v, q[la - 1].v)) < EPS) { 77 la--; 78 if (q[la] == L[i]) q[++la] = L[i]; 79 else if (onLeft(q[la], L[i].p)) q[la] = L[i]; 80 } 81 if (fi < la) p[la - 1] = dLineIntersect(q[la - 1], q[la]); 82 // cout << p[la - 1].x << ' ' << p[la - 1].y << endl; 83 // cout << fi << "=3=" << la << endl; 84 } 85 while (fi < la && !onLeft(q[fi], p[la - 1])) la--; 86 // cout << fi << ' ' << la << endl; 87 for (int i = fi; i <= la; i++) ans[ansEnd++] = q[i].id; 88 } 89 90 double B[N], A; 91 92 int main() { 93 // freopen("in", "r", stdin); 94 int n; 95 while (cin >> n) { 96 for (int i = 0; i < n; i++) scanf("%lf", &B[i]); 97 for (int i = 0; i < n; i++) { 98 scanf("%lf", &A); 99 rec[i] = DLine(Point(0.0, B[i]), Vec(1.0, A)); 100 rec[i].id = i + 1; 101 } 102 rec[n] = DLine(Point(0, 1), Vec(0.0, -1.0)); 103 rec[n].id = 0; 104 halfPlane(rec, n + 1); 105 cout << ansEnd - 1 << endl; 106 sort(ans, ans + ansEnd); 107 for (int i = 1; i < ansEnd; i++) { 108 if (i > 1) putchar(' '); 109 printf("%d", ans[i]); 110 } 111 puts(""); 112 } 113 return 0; 114 }
——written by Lyon
