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 $4  imes N$ 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=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 - (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=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;
}

const int maxn = 100000 + 10;
int n;
Point points[maxn * 4], ch[maxn * 4];

int main()
{
    //printf("%lf
", sqrt(2) * 4);
    while(scanf("%d", &n) == 1)
    {
        double x, y;
        int cnt = 0;
        for(int i = 0;i < n;i++)
        {
            scanf("%lf%lf", &x, &y);
            points[cnt].x = x+1, points[cnt++].y = y;
            points[cnt].x = x-1, points[cnt++].y = y;
            points[cnt].x = x, points[cnt++].y = y+1;
            points[cnt].x = x, points[cnt++].y = y-1;
        }
        int pcnt = ConvexHull(points, cnt, ch);
        double ans = 0;

        //sort(ch, ch + pcnt);

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

        double dx = fabs(ch[0].x - ch[pcnt-1].x);
        double dy = fabs(ch[0].y - ch[pcnt-1].y);
        if(dx > dy)  swap(dx, dy);
        ans += dx * sqrt(2);
        ans += (dy - dx);

        for(int i = 1;i < pcnt;i++)
        {
            double dx = fabs(ch[i].x - ch[i-1].x);
            double dy = fabs(ch[i].y - ch[i-1].y);
            if(dx > dy)  swap(dx, dy);
            ans += dx * sqrt(2);
            ans += (dy - dx);

            //printf("%lf
", ans);
        }
        printf("%lf
", ans);
    }
    return 0;
}

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