題目鏈接

分析：
如果固定了A不動，那么Q點則以P為圓心，PQ為半徑做圓周運動。然后B與Q的相對位置保持不變。那么B上的任意一個點D，則繞著點P＋QB，以PQ為半徑做圓周運動。題目變成了A是否和B上任意一個頂點畫的圓相交的問題了。判斷A與圓相交是板子了。同理也固定B不動，A做圓周運動判一下。

代碼：

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cstdlib> #include <vector> #include <queue> #include <deque> #include <stack> #include <set> #include <cmath>using namespace std;struct Point {double x, y;Point (double x = 0, double y = 0) : x(x), y(y) {} };struct Segment {Point A, B;Segment(Point a, Point b) : A(a), B(b) {}Segment(){} };typedef Point Vector;Vector operator +(Vector A, Vector B) {return Vector(A.x + B.x, A.y + B.y);} Vector operator -(Point A, Point B) {return Vector(A.x - B.x, A.y - B.y);} Vector operator *(Vector A, double p) {return Vector(A.x * p, A.y * p);} Vector operator /(Vector A, double p) {return Vector(A.x / p, A.y / p);} bool operator <(const Point &a, const Point &b) {return a.x < b.x || (a.x == b.x && a.y < b.y); }const double eps = 1e-10; int dcmp(double x) {if (fabs(x) < eps) return 0;else return x < 0 ? -1 : 1; }bool operator ==(const Point &a, const Point &b) {return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; }double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y; } double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x; } double Length(Vector A) {return sqrt(Dot(A, A)); } double DistanceToSegment(Point P, Point A, Point B) {if (A == B) return (Length(P - A));Vector v1 = B - A, v2 = P - A, v3 = P - B;if (dcmp(Dot(v1, v2)) < 0) return Length(v2);else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);else return fabs(Cross(v1, v2)) / Length(v1); } double mDistanceToSegment(Point P, Point A, Point B) {double ans = -1;double tmp = Length(P - A);ans = max(ans, tmp);tmp = Length(P - B);ans = max(ans, tmp);return ans; }const int maxn = 1e3 + 5;Segment va[maxn], vb[maxn]; Point pa[maxn], pb[maxn]; int na, nb; Point P, Q; double R;bool check(Point p, Point q, Segment *a, int n, Point *b, int m) {for (int i = 0; i < m; i ++) {Point o = p + (b[i] - q);for (int j = 0; j < n; j ++) {double minv = DistanceToSegment(o, a[j].A, a[j].B);double maxv = mDistanceToSegment(o, a[j].A, a[j].B);if (minv <= R && R <= maxv) return false;}}return true; }int main(int argc, char const *argv[]) {cin>>P.x>>P.y;cin>>na;for (int i = 0; i < na; i ++)cin>>pa[i].x>>pa[i].y;cin>>Q.x>>Q.y;cin>>nb;for (int i = 0; i < nb; i ++)cin>>pb[i].x>>pb[i].y;R = Length(P - Q);va[0] = Segment(pa[0], pa[na - 1]);for (int i = 1; i < na; i ++)va[i] = Segment(pa[i - 1], pa[i]);vb[0] = Segment(pb[0], pb[nb - 1]);for (int i = 1; i < nb; i ++)vb[i] = Segment(pb[i - 1], pb[i]);if (check(P, Q, va, na, pb, nb) && check(Q, P, vb, nb, pa, na)) puts("NO");else puts("YES");return 0; }

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