UVALive 6859——凸包&&周长

题目

链接

题意：在一个网格图上，给出$n$个点的坐标，用一个多边形包围这些点（不能接触，且多边形的边只能是对角线或直线），求多边形的最小周长.

分析

对于每个点，我们考虑与之相邻的4个点。一共由 $4  \times N$ 个点，然后求凸包。对凸包上每对相邻的点，优先走对角线，然后走直线。累加长度即可。

 #include<bits/stdc++.h>
 using namespace std;

 struct Point{
     double x, y;
     Point(double x=, double y=):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 - (Point A, Point 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);
 }
 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=;
     for(int i = ;i < n;i++)
     {
         while(m >  && Cross(ch[m-] - ch[m-], p[i] - ch[m-]) <= )  m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-;i >= ;i--)
     {
         while(m > k && Cross(ch[m-] - ch[m-], p[i] - ch[m-]) <= )  m--;
         ch[m++] = p[i];
     }
     if(n > )  m--;
     return m;
 }

 const int maxn =  + ;
 int n;
 Point points[maxn * ], ch[maxn * ];

 int main()
 {
     //printf("%lf\n", sqrt(2) * 4);
     while(scanf("%d", &n) == )
     {
         double x, y;
         int cnt = ;
         for(int i = ;i < n;i++)
         {
             scanf("%lf%lf", &x, &y);
             points[cnt].x = x+, points[cnt++].y = y;
             points[cnt].x = x-, points[cnt++].y = y;
             points[cnt].x = x, points[cnt++].y = y+;
             points[cnt].x = x, points[cnt++].y = y-;
         }
         int pcnt = ConvexHull(points, cnt, ch);
         double ans = ;

         //sort(ch, ch + pcnt);

         //for(int i = 0;i < pcnt;i++)  printf("%d: %lf  %lf\n", i, ch[i].x, ch[i].y);

         double dx = fabs(ch[].x - ch[pcnt-].x);
         double dy = fabs(ch[].y - ch[pcnt-].y);
         if(dx > dy)  swap(dx, dy);
         ans += dx * sqrt();
         ans += (dy - dx);

         for(int i = ;i < pcnt;i++)
         {
             double dx = fabs(ch[i].x - ch[i-].x);
             double dy = fabs(ch[i].y - ch[i-].y);
             if(dx > dy)  swap(dx, dy);
             ans += dx * sqrt();
             ans += (dy - dx);

             //printf("%lf\n", ans);
         }
         printf("%lf\n", ans);
     }
     return ;
 }

参考链接：https://blog.csdn.net/qingshui23/article/details/52809736