题意：顺时针方向给出N个点，求外围距离这些点L距离的点围成的图形的周长，结果四舍五入到整数(3 <= N <= 1000，1 <= L <= 1000，-10000 <= Xi, Yi <= 10000) 。
题目链接：http://poj.org/problem?id=1113
——>>先求凸包，然后求凸包的周长加上一个半径为L的圆的周长。
注意：用round()四舍五入后，若用%.0lf输出会WA，round()后强转为int后输出AC!
#include <cstdio>
#include <cmath>
#include <algorithm>using namespace std;const int maxn = 1000 + 10;
const double eps = 1e-10;
const double pi = acos(-1);int dcmp(double x){if(fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;
}struct Point{double x;double y;Point(double x = 0, double y = 0):x(x), y(y){}bool operator < (const Point& e) const{return x < e.x || (dcmp(x - e.x) == 0 && y < e.y);}
}p[maxn], q[maxn];typedef Point Vector;Vector operator + (Point A, Point 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 * (Point A, double p){return Vector(A.x * p, A.y * p);
}Vector operator / (Point A, double p){return Vector(A.x / p, A.y / p);
}double Cross(Vector A, Vector B){return A.x * B.y - B.x * A.y;
}int ConvexHull(Point *p, int n, Point* ch){ //求凸包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;
}double Dis(Point A, Point B){return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}int main()
{int N, L;while(scanf("%d%d", &N, &L) == 2){for(int i = 0; i < N; i++) scanf("%lf%lf", &p[i].x, &p[i].y);double ret = 0;int m = ConvexHull(p, N, q);for(int i = 1; i < m; i++) ret += Dis(q[i], q[i-1]);ret += Dis(q[0], q[m-1]) + 2 * pi * L;int ans = round(ret);printf("%d\n", ans);}return 0;
}