POJ 1113 wall (凸包）

来源：互联网发布：中密西根大学知乎编辑：程序博客网时间：2024/06/05 16:36
解题思路:
凸包的周长加上原的周长，最后四舍五入。
#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include <map>#define LL long long using namespace std;const int MAXN = 1000 + 10;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);}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 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; }int ConvexHull(Point* p, int n, Point* ch){sort(p, p+n);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;}//Point p[MAXN], ch[MAXN];double dis(Point A, Point B){double x = A.x - B.x;double y = A.y - B.y;return (double) sqrt(x *x  + y * y);}const double PI = acos(-1);int main(){int N , L;while(scanf("%d%d", &N, &L)!=EOF){Point p[MAXN], ch[MAXN];for(int i=0;i<N;i++)scanf("%lf%lf", &p[i].x, &p[i].y);int m = ConvexHull(p, N, ch);double ans = 0.0;for(int i=0;i<m;i++){if(i != m -1) ans += dis(ch[i], ch[i+1]);else ans += dis(ch[i], ch[0]);}ans += (double) 2 * PI * L;printf("%d\n", (int) (ans + 0.5));}return 0;}
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