hdu ?：?http://acm.hdu.edu.cn/showproblem.php?pid=1392? ?

題意：給你n個數的坐標，用一根繩子把它們圍起來，問你需要多長的繩子？

（當數的個數為2時特判 即兩點的距離，這是因為一條直線不應再回去圍了）

題解：簡單的凸包；

code：

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cmath> using namespace std; #define eps 1e-6 int top , n; //點 struct POINT {double x, y;POINT(){ }POINT(double a, double b){x = a;y = b;} }p[105],st[105]; //線段 struct Seg {POINT a, b;Seg() { }Seg(POINT x, POINT y){a = x;b = y;} }; //叉乘 double cross(POINT o, POINT a, POINT b) {return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y); }//多邊形面積，需要有順序，順(逆)時針。double area() {double ans = 0;for(int i = 1; i < top; i ++){ans += cross(p[0], p[i], p[i + 1]);}return ans; } //找凸包基點排序 bool cmp0(POINT a, POINT b) {if(a.y < b.y) return true;else if(a.y == b.y && a.x < b.x) return true;return false; } double dis(POINT a,POINT b){return sqrt( (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) ); } //極角排序 bool cmp1(POINT a, POINT b) {if(cross(p[0], a, b) > eps) return true;else if(fabs(cross(p[0], a, b)) < eps && dis(p[0], a) - dis(p[0], b) > eps) return true;return false; } //Graham_scan 求凸包.所求為純凈凸包... void Graham_scan() {sort(p, p + n, cmp0);sort(p + 1, p + n, cmp1);top = 0;p[n] = p[0];st[top ++] = p[0]; st[top ++] = p[1];for(int i = 2; i <= n; i ++){while(top > 2 && (cross(st[top - 1], st[top - 2], p[i]) > eps || fabs(cross(st[top - 1], st[top - 2], p[i])) < eps)) top --;st[top ++] = p[i];}top --; }int main(){while(scanf("%d",&n),n){for(int i = 0;i < n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);if(n == 2) {printf("%.2lf\n",dis(p[0],p[1]));continue;}Graham_scan();double sumd = 0;for(int i = 0;i < top;i++)sumd += dis(st[i],st[i+1]);printf("%.2lf\n",sumd);} }

poj ? :?http://poj.org/problem?id=2365??

題意：類似于上題，不過會多加一個圓的周長，因為繩子是繞圓柱的圍成多邊形后即多了一個圓的周長；

總結

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