hdu2202 最大三角形（旋转卡壳）

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hdu2202

题目

中文题

思路

学习了什么是旋转卡壳，求三角形面积的话就是多加了一个底边是哪两个点组成的。

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define pi acos (-1)#define rotate Rotateusing namespace std;typedef long long ll;const double eps = 1e-8;const int maxn=50010;const double INF = 1e20;int dcmp (double x){    if (fabs (x) < eps)        return 0;    else return x < 0 ? -1 : 1;}struct point{    double x, y;    point (double _x = 0, double _y = 0) : x(_x), y(_y) {}    point operator - (point a) const    {        return point (x-a.x, y-a.y);    }    point operator + (point a) const    {        return point (x+a.x, y+a.y);    }    bool operator < (const point &a) const    {        return x < a.x || (x == a.x && y < a.y);    }    bool operator == (const point &a) const    {        return dcmp (x-a.x) == 0 && dcmp (y-a.y) == 0;    }};point operator * (point a, double p){    return point (a.x*p, a.y*p);}double cross (point a, point b){    return a.x*b.y-a.y*b.x;}double dot (point a, point b){    return a.x*b.x + a.y*b.y;}int ConvexHull (point *p, point *ch, int n){    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-1]) <= 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-1]) <= 0)            m--;        ch[m++] = p[i];    }    if (n > 1)        m--;    return m;}int cnt,n;point p[maxn],ch[maxn];bool OnSegment (point p, point a1, point a2){    return dcmp (cross (a1-p, a2-p)) == 0 && dcmp (dot (a1-p, a2-p)) < 0;}double rotate_calipers (int m){    if (m == 1 || m == 2)        return 0;    double ans = 0;    int cur = 1;    for (int add = 1; add < m; add++)    {        for (int i = 0; i < m; i++)        {            while (cross (ch[i]-ch[(i+add)%m], ch[(cur+1)%m]-ch[cur]) < 0)                cur = (cur+1)%m;            double res = max (cross (ch[(i+add)%m]-ch[i], ch[cur]-ch[i]), cross (ch[(i+add)%m]-ch[i], ch[(cur+1)%m]-ch[i]));            ans = max (ans, res);        }    }    return ans/2.0;}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=0; i<n; i++)            scanf("%lf %lf",&p[i].x,&p[i].y);        cnt=ConvexHull(p,ch,n);        printf("%.2lf\n",rotate_calipers(cnt));    }    return 0;}

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