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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348
求凸包周长+2*PI*L
#include <stdio.h> #include <algorithm> #include <cstring> #include <cmath> using namespace std; const int N = 110; const double eps = 1e-6; const double PI = acos(-1); struct point { double x, y; point(){} point(double x, double y) : x(x), y(y) {} point friend operator - (const point &p1, const point &p2)///矢量p2p1; { return point(p1.x-p2.x, p1.y-p2.y); } double friend operator ^ (const point &p1, const point &p2)///p1×p2; { return p1.x*p2.y - p1.y*p2.x; } }; point p[N], res[N]; double Dist(point p1, point p2) { return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } int cmp1(point p1, point p2)///位置排序,找到最下方的; { if(p1.y != p2.y) return p1.y < p2.y; return p1.x < p2.x;///若有多个下方的找左边的; } int cmp2(point p1, point p2)///极角排序;若极角相同,距离近的在前面; { double k = (p1-p[0])^(p2-p[0]); if( k>eps || (fabs(k)<eps && Dist(p1, p[0]) < Dist(p2, p[0]) )) return 1; return 0; } int Graham(int n)///构造凸包,O(nlogn) { res[0] = p[0]; res[1] = p[1]; res[2] = p[2]; int top = 2; for(int i=3; i<n; i++) { while(((p[i]-res[top])^(res[top-1]-res[top])) <= 0)top--; ///当res[top-1]->res[top]->p[i],出现右拐或直行时没说明res[top]不是凸包上的点; res[++top] = p[i]; } return top; } int main() { int n, T; double L; scanf("%d", &T); while(T--) { scanf("%d%lf", &n, &L); for(int i=0; i<n; i++) scanf("%lf %lf", &p[i].x, &p[i].y); sort(p, p+n, cmp1);///p[0]为最下方靠左的点; sort(p+1, p+n, cmp2);///以p[0]为基点,按叉积进行排序; int cnt = Graham(n);///求凸包的顶点个数cnt+1,保存在res中,下标从0开始; double ans = Dist(res[0], res[cnt]); for(int i=1; i<=cnt; i++) ans += Dist(res[i], res[i-1]); ans += PI*2*L; printf("%.0f\n", ans); if(T)puts(""); } return 0; }
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原文地址:http://www.cnblogs.com/zhengguiping--9876/p/5908588.html