旋轉卡殼 半平面交板子 poj 2079

sampleinput

編輯

3

3 42 6
2 75
2 63 9
2 08 0
6 5-1

sampleoutput

編輯

0.50

27.00

#pragma comment(linker, "/stack:1024000000,1024000000")

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-8
using namespace std;
const int maxn=2e3+5;
int m;
double r;
int ccnt,curcnt;//此時ccnt為最終切割得到的多邊形的頂點數、暫存頂點個數
struct point
point(double _x,double _y):x(_x),y(_y){}
void input()
};point points[maxn],p[maxn],q[maxn];//讀入的多邊形的頂點（順時針）、p為存放最終切割得到的多邊形頂點的陣列、暫存核的頂點
void getline(point x,point y,double &a,double &b,double &c)
void initial()
point intersect(point x,point y,double a,double b,double c)
void cut(double a,double b,double c)
if(a*p[i+1].x + b*p[i+1].y + c > 0)}}
for(int i = 1; i <= curcnt; ++i)p[i] = q[i];//將q中暫存的核的頂點轉移到p中
p[curcnt+1] = q[1];p[0] = p[curcnt];
ccnt = curcnt;
}void solve()
/*如果要向內推進r，用該部分代替上個函式
for(int i = 1; i <= m; ++i)
for(int i = 1; i <= ccnt; ++i)}}
printf("%.4lf %.4lf %.4lf %.4lf/n",p[ansx].x,p[ansx].y,p[ansy].x,p[ansy].y); */
//多邊形核的面積
double area = 0;
for(int i = 1; i <= curcnt; ++i)
area += p[i].x * p[i + 1].y - p[i + 1].x * p[i].y;
area = fabs(area / 2.0);
printf("%.2f\n",area);
}inline void guizhenghua()
inline void init()
int main()
}

//china no.1

#pragma comment(linker, "/stack:1024000000,1024000000")
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;
templateinline t max(t a,t b,t c) 
#define eps 1e-12
#define ll long long
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define feq(a,b) (fabs((a)-(b))eps+(b)) //a > b
#define fsm(a,b) ((b) > eps+(a)) //a < b
#define fbgeq(a,b) ((a) >= (b)-eps) //a >= b
#define fsmeq(a,b) ((b) >= (a)-eps) //a <= b
struct point
point(double _x,double _y):x(_x),y(_y){}
point operator -(const point &b)const
int operator ^(const point &b)const
int operator *(const point &b)const
void input()
};const int maxn = 100110;
inline double pdis(point a,point b)
inline double difcross(point p0,point p1,point p2)
inline double dis(point a,point b)
inline double rotate_caliper(point p,int n)
}*///旋轉卡殼，求兩點間距離平方的最大值
point v;
for(int i = 0;i < n;i++)
//return ans/2.0;
return ans;
}point p0;
int graham[maxn],top,n;
point points[maxn],p[maxn];
inline int cmp(point a, point b)
inline void convexhull()
swap(points[0],points[u]);
p0 = points[0];
sort( points+1, points + n, cmp );
if(n <= 2)
return;
}graham[0] = 0;
graham[1] = 1;
graham[2] = 2;
top = 2;
for( i = 3; i < n; i++)
graham[++top] = i;
}top ++;
for(int i = 0;i < top; ++i)
p[i] = points[graham[i]];
}int main()
*/ convexhull();
double ans = rotate_caliper(p,top);
printf("%lld\n",(ll)ans);
}return 0;
}

sort(points,points+n,point_cmp);

n=unique(points,points+n)-points;
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