题意：求凸包面积/50，并取整。
分析：用模板。
View Code #include#include #include #include #include using namespace std;#define maxn 10005struct Point{ double x, y;} point[maxn], cvx[maxn];int n, m;bool mult(Point sp, Point ep, Point op){ return (sp.x - ep.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - ep.y);}bool operator <(const Point &l, const Point &r){ return l.y < r.y || (l.y == r.y && l.x < r.x);}int graham(Point pnt[], int n, Point res[]){ int i, len, top = 1; sort(pnt, pnt + n); if (n == 0) return 0; res[0] = pnt[0]; if (n == 1) return 1; res[1] = pnt[1]; if (n == 2) return 2; res[2] = pnt[2]; for (i = 2; i < n; i++) { while (top && mult(pnt[i], res[top], res[top - 1])) top--; res[++top] = pnt[i]; } len = top; res[++top] = pnt[n - 2]; for (i = n - 3; i >= 0; i--) { while (top != len && mult(pnt[i], res[top], res[top - 1])) top--; res[++top] = pnt[i]; } return top; // 返回凸包中点的个数}void input(){ scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%lf %lf", &point[i].x, &point[i].y);}double areaofp(int vcount, Point plg[]){ int i; double s; if (vcount < 3) return 0; s = plg[0].y * (plg[vcount - 1].x - plg[1].x); for (i = 1; i < vcount; i++) s += plg[i].y * (plg[(i - 1)].x - plg[(i + 1) % vcount].x); return s / 2;}int main(){ //freopen("t.txt", "r", stdin); input(); m = graham(point, n, cvx); printf("%d\n", (int)(areaofp(m, cvx) / 50)); return 0;}