hdu 2202 最大三角形 (Convex Hull && (Bruce Force || Rotate Stuck))
题目中文，不另外解释。
做法是构造出凸包，估计点是随机出的，所以凸包的大小不会太大，于是可以直接对凸包上的点进行暴力计算三角形的面积。
如果随机的点是n个，凸包上面有m个点，那么复杂度就是O(n+m^3)。代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 #include <iostream> 6 7 using namespace std; 8 9 const double EPS = 1e-8; 10 inline int sgn(double x) { return (x > EPS) - (x < -EPS);} 11 struct Point { 12 double x, y; 13 Point() {} 14 Point(double x, double y) : x(x), y(y) {} 15 bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && y < a.y;} 16 bool operator == (Point a) const { return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;} 17 Point operator + (Point a) const { return Point(x + a.x, y + a.y);} 18 Point operator - (Point a) const { return Point(x - a.x, y - a.y);} 19 Point operator * (double p) const { return Point(x * p, y * p);} 20 Point operator / (double p) const { return Point(x / p, y / p);} 21 } ; 22 typedef Point Vec; 23 inline double crossDet(Point a, Point b) { return a.x * b.y - a.y * b.x;} 24 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);} 25 inline double dotDet(Point a, Point b) { return a.x * b.x + a.y * b.y;} 26 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));} 27 28 struct Line { 29 Point s, t; 30 Line() {} 31 Line(Point s, Point t) : s(s), t(t) {} 32 Vec vec() { return t - s;} 33 Point point(double x) { return s + (t - s) * x;} 34 } ; 35 typedef Line Seg; 36 double pt2Line(Point x, Point a, Point b) { 37 Vec v1 = b - a, v2 = x - a; 38 return crossDet(v1, v2) / vecLen(v1); 39 } 40 inline double pt2Line(Point x, Line L) { return pt2Line(x, L.s, L.t);} 41 42 int andrew(Point *pt, int n, Point *ch) { 43 sort(pt, pt + n); 44 int m = 0; 45 for (int i = 0; i < n; i++) { 46 while (m > 1 && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--; 47 ch[m++] = pt[i]; 48 } 49 int k = m; 50 for (int i = n - 2; i >= 0; i--) { 51 while (m > k && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--; 52 ch[m++] = pt[i]; 53 } 54 if (n > 1) m--; 55 return m; 56 } 57 58 const int N = 55555; 59 const double FINF = 1e20; 60 Point pt[N], ch[N]; 61 62 int main() { 63 // freopen("in", "r", stdin); 64 int n; 65 while (cin >> n) { 66 for (int i = 0; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y); 67 n = andrew(pt, n, ch); 68 // for (int i = 0; i < n; i++) { 69 // cout << ch[i].x << ' ' << ch[i].y << endl; 70 // } 71 double maxArea = 0.0; 72 for (int i = 0; i < n; i++) { 73 for (int j = i + 1; j < n; j++) { 74 for (int k = j + 1; k < n; k++) { 75 maxArea = max(maxArea, fabs(crossDet(ch[i], ch[j], ch[k])) / 2.0); 76 } 77 } 78 } 79 printf("%.2f\n", maxArea); 80 } 81 return 0; 82 }
还有一种做法，就是O(n+m^2)求对踵点的方法。每次枚举其中一条边，然后扫描求出对踵点，移动一条边的一个端点，同时继续搜索对踵点。这里对于每一条新的边，对踵点都是继续往下一个点移动。这样子，对于每个确定的端点都是只需要访问O(m)个对踵点。这样的操作共有m次，所以这里的复杂度是O(m^2)。代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 #include <iostream> 6 7 using namespace std; 8 9 const double EPS = 1e-8; 10 inline int sgn(double x) { return (x > EPS) - (x < -EPS);} 11 struct Point { 12 double x, y; 13 Point() {} 14 Point(double x, double y) : x(x), y(y) {} 15 bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && y < a.y;} 16 bool operator == (Point a) const { return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;} 17 Point operator + (Point a) const { return Point(x + a.x, y + a.y);} 18 Point operator - (Point a) const { return Point(x - a.x, y - a.y);} 19 Point operator * (double p) const { return Point(x * p, y * p);} 20 Point operator / (double p) const { return Point(x / p, y / p);} 21 } ; 22 typedef Point Vec; 23 inline double crossDet(Point a, Point b) { return a.x * b.y - a.y * b.x;} 24 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);} 25 inline double dotDet(Point a, Point b) { return a.x * b.x + a.y * b.y;} 26 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));} 27 28 struct Line { 29 Point s, t; 30 Line() {} 31 Line(Point s, Point t) : s(s), t(t) {} 32 Vec vec() { return t - s;} 33 Point point(double x) { return s + (t - s) * x;} 34 } ; 35 typedef Line Seg; 36 double pt2Line(Point x, Point a, Point b) { 37 Vec v1 = b - a, v2 = x - a; 38 return crossDet(v1, v2) / vecLen(v1); 39 } 40 inline double pt2Line(Point x, Line L) { return pt2Line(x, L.s, L.t);} 41 42 int andrew(Point *pt, int n, Point *ch) { 43 sort(pt, pt + n); 44 int m = 0; 45 for (int i = 0; i < n; i++) { 46 while (m > 1 && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--; 47 ch[m++] = pt[i]; 48 } 49 int k = m; 50 for (int i = n - 2; i >= 0; i--) { 51 while (m > k && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--; 52 ch[m++] = pt[i]; 53 } 54 if (n > 1) m--; 55 return m; 56 } 57 58 const int N = 55555; 59 const double FINF = 1e20; 60 Point pt[N], ch[N]; 61 62 int main() { 63 // freopen("in", "r", stdin); 64 int n; 65 while (cin >> n) { 66 for (int i = 0; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y); 67 n = andrew(pt, n, ch); 68 ch[n] = ch[0]; 69 // for (int i = 0; i < n; i++) { 70 // cout << ch[i].x << ' ' << ch[i].y << endl; 71 // } 72 double maxArea = 0.0; 73 for (int i = 0; i < n; i++) { 74 int mk = i; 75 double last = 0.0, cur = fabs(pt2Line(ch[mk], ch[i], ch[i + 1])); 76 while (true) { 77 if (sgn(last - cur) > 0) break; 78 last = cur; 79 mk = (mk + 1) % n; 80 cur = pt2Line(ch[mk], ch[i], ch[i + 1]); 81 } 82 mk = (mk + n - 1) % n; 83 maxArea = max(maxArea, fabs(crossDet(ch[mk], ch[i], ch[i + 1])) / 2.0); 84 for (int j = 2; j < n - 1; j++) { 85 int k = (i + j) % n; 86 last = 0.0, cur = fabs(pt2Line(ch[mk], ch[i], ch[k])); 87 while (true) { 88 if (sgn(last - cur) > 0) break; 89 last = cur; 90 mk = (mk + 1) % n; 91 cur = pt2Line(ch[mk], ch[i], ch[k]); 92 } 93 mk = (mk + n - 1) % n; 94 maxArea = max(maxArea, fabs(crossDet(ch[mk], ch[i], ch[k])) / 2.0); 95 } 96 } 97 printf("%.2f\n", maxArea); 98 } 99 return 0; 100 }
对于最特殊的情况，也就是有50000个点在凸包上的算法暂时没有想到。
——written by Lyon
