题目链接:点击打开链接
思路:首先,我们要知道一个贪心结论:敌人如果有k个炸弹, 那么他一定是炸连续的k个点, 这样会使得炸的面积最大。 那么我们只要二分炸弹数mid,每隔mid个点重新建立一个平面, 仍然是n个平面, 代表n种可能情况, 那么如果他们的交存在, 那么司令部只要放在这个平面交的面积内就行了。 所以问题迎刃而解, 二分答案, 用半平面交判断答案是否可行。
细节参见代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include <cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) using namespace std; typedef long long ll; typedef long double ld; const double eps = 1e-6; const double PI = acos(-1); const int mod = 1000000000 + 7; const int INF = 0x3f3f3f3f; const int seed = 131; const ll INF64 = ll(1e18); const int maxn = 50000 + 10; int T,n,m; struct point { double x, y; point(double x=0, double y=0):x(x), y(y) {} }; typedef point Vector; Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); } Vector operator - (Vector A, Vector 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); } 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 cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } double Dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } Vector Normal(Vector A) { double L = Length(A); return Vector(-A.y/L, A.x/L); } struct Line { point p; Vector v; double ang; Line() {} Line(point p, Vector v) : p(p), v(v) { ang = atan2(v.y, v.x); } bool operator < (const Line& L) const { return ang < L.ang; } }; bool onLeft(Line L, point p) { return cross(L.v, p-L.p) > 0; } point getintersection(Line a, Line b) { Vector u = a.p - b.p; double t = cross(b.v, u) / cross(a.v, b.v); return a.p + a.v * t; } int halfplane(Line* L, int n, point* poly) { sort(L, L+n); int first, last; point *p = new point[n]; Line *q = new Line[n]; q[first=last=0] = L[0]; for(int i = 1; i < n; i++) { while(first < last && !onLeft(L[i], p[last-1])) last--; while(first < last && !onLeft(L[i], p[first])) first++; q[++last] = L[i]; if(fabs(cross(q[last].v, q[last-1].v)) < eps) { last--; if(onLeft(q[last], L[i].p)) q[last] = L[i]; } if(first < last) p[last-1] = getintersection(q[last-1], q[last]); } while(first < last && !onLeft(q[first], p[last-1])) last--; if(last - first <= 1) return 0; p[last] = getintersection(q[last], q[first]); int m = 0; for(int i = first; i <= last; i++) poly[m++] = p[i]; return m; } point p[maxn], poly[maxn]; Line L[maxn]; double x, y; int main() { while(~scanf("%d",&n) && n) { for(int i = 0; i < n; i++) { scanf("%lf%lf",&x,&y); p[i] = point(x, y); } int l = 1, r = n - 1; while(r > l) { int mid = (r + l) / 2; for(int i = 0; i < n; i++) L[i] = Line(p[i], p[i] - p[(i+mid+1)%n]); int m = halfplane(L, n, poly); if(!m) r = mid; else l = mid+1; } printf("%d\n", l); } return 0; }
时间: 2024-10-29 16:19:42