[计算几何笔记2]旋转卡壳

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POJ 2187

大概旋转卡壳就是这个意思了。。

int rotating_calipers(Point *ch, int m) /**旋转卡壳模板*/{    int q = 1;    int ans = 0;    ch[m] = ch[0]; /**凸包边界处理*/    for(int i = 0; i < m; i++) /**依次用叉积找出凸包每一条边对应的最高点*/    {/**同底不同高,每次用面积判断高的大小就可以了*/        while(Cross(ch[i+1]-ch[i], ch[q+1]-ch[i]) > Cross(ch[i+1]-ch[i], ch[q]-ch[i]))            q = (q+1)%m;    /**每条底上有两个点,所以要 max 两次*/        ans = max(ans, max(squarDist(ch[i], ch[q]), squarDist(ch[i+1], ch[q+1])));    }    return ans;/**返回的也就是凸包的直径*/}

另附POJ2187

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define maxn 200010using namespace std;//-------------------------------------------------------------//int n;struct Point{int x, y;bool operator<(const Point& k)const{if(x != k.x)return x < k.x;return y < k.y;}bool operator==(const Point& k)const{return x == k.x && y == k.y;}Point(int x = 0, int y = 0):x(x), y(y){}void read(){scanf("%d%d", &x, &y);}void print(){printf("%d %d\n", x, y);}}b[maxn], st[maxn], a[maxn];Point operator-(const Point& a, const Point& b){return Point(a.x - b.x, a.y - b.y);}int Cross(const Point& a, const Point& b){return a.x * b.y - a.y * b.x;}int length(const Point& k){return k.x * k.x + k.y * k.y;}const double eps = 1e-8;int Filter(double x){return x > eps ? 1 : (x < -eps ? -1 : 0);}//-------------------------------------------------------------//int main(){scanf("%d", &n);for(int i = 1; i <= n;i ++)    b[i].read();sort(b+1, b+1+n);int top = 0;st[++ top] = b[1];st[++ top] = b[2];for(int i = 3; i <= n; i ++){while(top > 1 && Cross(st[top] - st[top - 1], b[i] - st[top - 1]) <= 0)    top --;st[++ top] = b[i];}for(int i = 1; i <= top; i ++)a[i] = st[i];int cnt = top;top = 0;st[++ top] = b[n];st[++ top] = b[n - 1];for(int i = n - 2; i; i --){while(top > 1 && Cross(st[top] - st[top - 1], b[i] - st[top - 1]) <= 0)    top --;st[++ top] = b[i];}for(int i = 1; i <= top; i ++){if(a[cnt] == st[i])continue;a[++ cnt] = st[i];}int ans = 0;int r = 2;a[cnt+1] = a[2];for(int i = 1; i < cnt; i ++){while(Cross(a[i+1] - a[i], a[r+1] - a[i]) > Cross(a[i+1] - a[i], a[r] - a[i]))r = r % cnt + 1;ans = max(length(a[r] - a[i]), ans);ans = max(length(a[r+1] - a[i+1]), ans);}printf("%d\n", ans);return 0;}

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