http://poj.org/problem?id=3348
水。
完整代码:
/*0ms,680KB*/
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
const int mx = 10005;
struct point
{
double x, y;
point(double x = 0.0, double y = 0.0): x(x), y(y) {}
void read() const
{
scanf("%lf%lf", &x, &y);
}
bool operator == (const point& a) const
{
return fabs(x - a.x) < eps && fabs(y - a.y) < eps;
}
bool operator < (const point& a) const
{
return x < a.x || fabs(x - a.x) < eps && y < a.y;
}
};
point a[mx], ans[mx];
int n, len;
///oa x ob
inline double cross_product(point& a, point& b)
{
return a.x * b.y - a.y * b.x;
}
///ab x ac
inline double cross_product(point& a, point& b, point& c)
{
return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
///求凸包
void convex_hull()
{
sort(a, a + n);
//unique(a, a + n);
len = 0;
int i;
for (i = 0; i < n; ++i)
{
while (len >= 2 && cross_product(ans[len - 2], ans[len - 1], a[i]) < eps)
--len;
ans[len++] = a[i];
}
int tmp = len;
for (i = n - 2; i >= 0; --i)
{
while (len > tmp && cross_product(ans[len - 2], ans[len - 1], a[i]) < eps)
--len;
ans[len++] = a[i];
}
--len;
}
///求多边形面积 的两倍 。以原点为向量起点。注意ans[]要是顺时针顺序
double area()
{
double sum = 0.0;
ans[len] = ans[0];
for (int i = 0; i < len; ++i) sum += cross_product(ans[i], ans[i + 1]);
return sum;
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; ++i) a[i].read();
if (n < 3) puts("0");
else
{
convex_hull();
printf("%d\n", (int)(area() / 100.0));
}
return 0;
}