題意：
給n個平面上的點，現在要在離點外l公尺處建城牆，使得城牆把所有點都包含進去且城牆的長度最短。
解析：韜哥出的某次訓練賽上a出的第一道計算幾何，算是大水題吧。
用convexhull演算法把凸包求出來，然後加加減減就a了。
計算見下圖：
好久沒玩畫圖了啊好開心。
**：
#include #include #include #include #include #include #include #include #include #include #define ll long long
using namespace std;
const int maxn = 1e3 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-10;
const double pi = 4 * atan(1.0);
struct point
};typedef point vector;
vector operator + (vector a, vector b)
vector operator - (vector a, vector b)
vector operator * (vector a, double p)
vector operator / (vector a, double p)
bool operator < (const point& a, const point& b)
int dcmp(double x)
else
}bool operator == (const point& a, const point& b)
double dot(vector a, vector b)
double length(vector a)
double angle(vector a, vector b)
double cross(vector a, vector b)
double area2(point a, point b, point c)
bool cmp(point a, point b)
int convexhull(point* p, int n, point* ch)
int k = m;
for (int i = n - 2; i >= 0; i--)
if (n > 1)
m--;
return m;
}int main()
point ch[maxn];
int index = convexhull(p, n, ch);
// for (int i = 0; i < index; i++)
// cout << ch[i].x << " " << ch[i].y << endl;
vector g[maxn];
double angle = 0;
double ans = 0;
for (int i = 0; i < index - 1; i++)
g[index - 1].x = ch[0].x - ch[index - 1].x;
g[index - 1].y = ch[0].y - ch[index - 1].y;
ans += length(g[index - 1]);
//cout << ans << endl;
for (int i = 0; i < index - 1; i++)
angle += angle(g[index - 1], g[0]);
//cout << angle << endl;
ans += pi * l * angle / pi;
printf("%.lf\n", ans);
}return 0;
}
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