UVA 1453 Squares

大意略。

思路：十万个点，如果直接枚举顶点的话会超时。可以求出凸包后，然后再枚举两个顶点，极限情况下也是O(n^2)，竟然被我水过去了。

正解是通过旋转卡壳的方法求凸包顶点之间的最小距离。

先贴暴力代码，然后去学习旋转卡壳：http://cgm.cs.mcgill.ca/~orm/rotcal.frame.html

#include <iostream>#include <cstdlib>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <stack>#include <algorithm>using namespace std;const double eps = 1e-10;const double PI = acos(-1.0);struct Point{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }bool operator < (const Point& a) const{if(a.x != x) return x < a.x;return y < a.y;}};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); }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 Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }double Area2(Point A, Point B, Point C) { return fabs(Cross(B-A, C-A)) / 2; }Vector Rotate(Vector A, double rad){return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){Vector u = P-Q;double t = Cross(w, u) / Cross(v, w);return P+v*t;}bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}bool OnSegment(Point p, Point a1, Point a2){return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}double PolygonArea(Point* p, int n){double area = 0;for(int i = 1; i < n-1; i++)area += Cross(p[i]-p[0], p[i+1]-p[0]);return area/2;}double PointDistanceToLine(Point P, Point A, Point B){Vector v1 = B-A, v2 = P-A;return fabs(Cross(v1, v2)) / Length(v1);}double PointDistanceToSegment(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);}int isPointInPolygon(Point p, Point *poly, int n){int wn = 0;for(int i = 0; i < n; i++){const Point& p1 = poly[i], p2 = poly[(i+1)%n];if(p == p1 || p == p2 || OnSegment(p, p1, p2)) return -1;int k = dcmp(Cross(p2-p1, p-p1));int d1 = dcmp(p1.y - p.y);int d2 = dcmp(p2.y - p.y);if(k > 0 && d1 <= 0 && d2 > 0) wn++;if(k < 0 && d2 <= 0 && d1 > 0) wn--;}if(wn != 0) return 1;return 0;}int ConvexHull(Point *p, int n, Point *ch) //凸包{sort(p, p+n);//n = unique(p, p+n) - p; //去重int m = 0;for(int i = 0; i < n; i++){while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;ch[m++] = p[i];}int k = m;for(int i = n-2; i >= 0; i--){while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;ch[m++] = p[i];}if(n > 1) m--;return m;}double PointToPoint(Point p1, Point p2){return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);}Point read_point(){Point A;scanf("%lf%lf", &A.x, &A.y);return A;}const int maxn = 100010;const double INF = 0x3f3f3f3f;Point P[maxn*4], Q[maxn*4];int n, pc;double x, y, w;void init(){pc = 0;}void read_case(){init();scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%lf%lf%lf", &x, &y, &w);P[pc++] = Point(x, y);P[pc++] = Point(x+w, y);P[pc++] = Point(x, y+w);P[pc++] = Point(x+w, y+w);}}void solve(){read_case();int m = ConvexHull(P, pc, Q);double ans = -INF, Max = -INF;for(int i = 0; i < m; i++){for(int j = i+1; j < m; j++){Max = PointToPoint(Q[i], Q[j]);ans = max(ans, Max);}}printf("%.0lf\n", ans);}int main(){int T;scanf("%d", &T);while(T--){solve();}return 0;}

 用旋转卡壳求点基直径：

#include <iostream>#include <cstdlib>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <stack>#include <algorithm>using namespace std;const double eps = 1e-10;const double PI = acos(-1.0);struct Point{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }bool operator < (const Point& a) const{if(a.x != x) return x < a.x;return y < a.y;}};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); }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 Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }double Area2(Point A, Point B, Point C) { return fabs(Cross(B-A, C-A)) / 2; }Vector Rotate(Vector A, double rad){return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){Vector u = P-Q;double t = Cross(w, u) / Cross(v, w);return P+v*t;}bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}bool OnSegment(Point p, Point a1, Point a2){return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}double PolygonArea(Point* p, int n){double area = 0;for(int i = 1; i < n-1; i++)area += Cross(p[i]-p[0], p[i+1]-p[0]);return area/2;}double PointDistanceToLine(Point P, Point A, Point B){Vector v1 = B-A, v2 = P-A;return fabs(Cross(v1, v2)) / Length(v1);}double PointDistanceToSegment(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);}int isPointInPolygon(Point p, Point *poly, int n){int wn = 0;for(int i = 0; i < n; i++){const Point& p1 = poly[i], p2 = poly[(i+1)%n];if(p == p1 || p == p2 || OnSegment(p, p1, p2)) return -1;int k = dcmp(Cross(p2-p1, p-p1));int d1 = dcmp(p1.y - p.y);int d2 = dcmp(p2.y - p.y);if(k > 0 && d1 <= 0 && d2 > 0) wn++;if(k < 0 && d2 <= 0 && d1 > 0) wn--;}if(wn != 0) return 1;return 0;}int ConvexHull(Point *p, int n, Point *ch) //凸包{sort(p, p+n);//n = unique(p, p+n) - p; //去重int m = 0;for(int i = 0; i < n; i++){while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;ch[m++] = p[i];}int k = m;for(int i = n-2; i >= 0; i--){while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;ch[m++] = p[i];}if(n > 1) m--;return m;}double Dist2(Point p1, Point p2){return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);}Point read_point(){Point A;scanf("%lf%lf", &A.x, &A.y);return A;}const int maxn = 100010;const double INF = 0x3f3f3f3f;Point P[maxn*4], Q[maxn*4];int n, pc;double x, y, w;void init(){pc = 0;}void read_case(){init();scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%lf%lf%lf", &x, &y, &w);P[pc++] = Point(x, y);P[pc++] = Point(x+w, y);P[pc++] = Point(x, y+w);P[pc++] = Point(x+w, y+w);}}// 返回点集直径的平方double RotatingCalipers(Point *P, int n) //旋转卡壳 {  if(n == 1) return 0;  if(n == 2) return Dist2(P[0], P[1]);  P[n] = P[0]; //避免取模  double ans = 0;  for(int u = 0, v = 1; u < n; u++)  {//一条直线贴住边p[u]-p[u+1]    for(;;){  // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转      // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0      // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)      // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0      double diff = Cross(P[u+1]-P[u], P[v+1]-P[v]);      if(diff <= 0)      {        ans = max(ans, Dist2(P[u], P[v])); // u和v是对踵点        if(diff == 0) ans = max(ans, Dist2(P[u], P[v+1])); // diff == 0时u和v+1也是对踵点        break;      }      v = (v + 1) % n;    }  }  return ans;}void solve(){read_case();int m = ConvexHull(P, pc, Q);double ans = RotatingCalipers(Q, m);printf("%.0lf\n", ans);}int main(){int T;scanf("%d", &T);while(T--){solve();}return 0;}