POJ 3348 Cows（凸包+多边形面积）

Description

Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.

However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.

Input

The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and y separated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).

Output

You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.

题目大意：给n个点，求凸包，然后求这个凸包的面积。

思路：跟题目大意一样……

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAXN = 10010;
const double EPS = 1e-8;
const double PI = acos(-1.0);//3.14159265358979323846

inline int sgn(double x) {
    return (x > EPS) - (x < -EPS);
}

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y): x(x), y(y) {}
    void read() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator == (const Point &rhs) const {
        return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
    }
    bool operator < (const Point &rhs) const {
        if(y != rhs.y) return y < rhs.y;
        return x < rhs.x;
    }
    Point operator + (const Point &rhs) const {
        return Point(x + rhs.x, y + rhs.y);
    }
    Point operator - (const Point &rhs) const {
        return Point(x - rhs.x, y - rhs.y);
    }
    Point operator * (const int &b) const {
        return Point(x * b, y * b);
    }
    Point operator / (const int &b) const {
        return Point(x / b, y / b);
    }
    double length() const {
        return sqrt(x * x + y * y);
    }
    Point unit() const {
        return *this / length();
    }
};
typedef Point Vector;

double dist(const Point &a, const Point &b) {
    return (a - b).length();
}

double cross(const Point &a, const Point &b) {
    return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
    return sgn(cross(sp - op, ed - op));
}

double area(const Point& a, const Point &b, const Point &c) {
    return fabs(cross(a - c, b - c)) / 2;
}

struct Seg {
    Point st, ed;
    Seg() {}
    Seg(Point st, Point ed): st(st), ed(ed) {}
    void read() {
        st.read(); ed.read();
    }
};
typedef Seg Line;

bool isOnSeg(const Seg &s, const Point &p) {
    return (p == s.st || p == s.ed) ||
        (((p.x - s.st.x) * (p.x - s.ed.x) < 0 ||
          (p.y - s.st.y) * (p.y - s.ed.y) < 0) &&
         sgn(cross(s.ed, p, s.st) == 0));
}

bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
    return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
        (max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
        (max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
        (max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
        (cross(s2, e1, s1) * cross(e1, e2, s1) >= 0) &&
        (cross(s1, e2, s2) * cross(e2, e1, s2) >= 0);
}

bool isIntersected(const Seg &a, const Seg &b) {
    return isIntersected(a.st, a.ed, b.st, b.ed);
}

bool isParallel(const Seg &a, const Seg &b) {
    return sgn(cross(a.ed - a.st, b.ed - b.st)) == 0;
}

//return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
    A = L.ed.y - L.st.y;
    B = L.st.x - L.ed.x;
    C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}

Point intersection(const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    Point I;
    I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
    I.y =   (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
    return I;
}

bool isEqual(const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    return sgn(A1 * B2 - A2 * B1) == 0 && sgn(A1 * C2 - A2 * C1) == 0 && sgn(B1 * C2 - B2 * C1) == 0;
}

struct Poly {
    int n;
    Point p[MAXN];//p[n] = p[0]
    void init(Point *pp, int nn) {
        n = nn;
        for(int i = 0; i < n; ++i) p[i] = pp[i];
        p[n] = p[0];
    }
    double area() {
        if(n < 3) return 0;
        double s = p[0].y * (p[n - 1].x - p[1].x);
        for(int i = 1; i < n; ++i)
            s += p[i].y * (p[i - 1].x - p[i + 1].x);
        return s / 2;
    }
};

void Graham_scan(Point *p, int n, int *stk, int &top) {//stk[0] = stk[top]
    sort(p, p + n);
    top = 1;
    stk[0] = 0; stk[1] = 1;
    for(int i = 2; i < n; ++i) {
        while(top && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top;
        stk[++top] = i;
    }
    int len = top;
    stk[++top] = n - 2;
    for(int i = n - 3; i >= 0; --i) {
        while(top != len && cross(p[i], p[stk[top]], p[stk[top - 1]]) >= 0) --top;
        stk[++top] = i;
    }
}

/*******************************************************************************************/

Point p[MAXN];
Poly poly;
int stk[MAXN], top;
int n, T;

int solve() {
    poly.n = top;
    for(int i = 0; i <= top; ++i) poly.p[i] = p[stk[i]];
    double ret = poly.area() + EPS;
    return int(ret / 50);
}

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) p[i].read();
    Graham_scan(p, n, stk, top);
    printf("%d
", solve());
}