Link Title: 1127. polygon area (Computational Geometry)
The meaning of problems
Given counterclockwise order \ (n-\) polygon area coordinates points, those points enclosed.
Thinking
Selecting a point on the polygon, and two points each after enumeration, the cross product is calculated, to retain the note symbol, the area of the polygon is the sum of all the results of cross-product.
For chestnut:
Computationally FIG polygons \ (ABCDEFGH \) area, select \ (A \) point, the area is equal to \ (\ frac {1} { 2} (\ boldsymbol {AB \ times AC} + \ boldsymbol {AC \ times AD +} \ {boldsymbol the AD \ Times the AE +} \ {boldsymbol the AE \ Times the AF} + \ {boldsymbol the AF \} + AG Times \ {boldsymbol AG \ Times the AH}) \) . Wherein \ (\ triangle ABC \) of the area is negative, the \ (\ triangle ACD \) and \ (\ triangle ADE \) area is positive, the polygon \ (ABCDE \) of the polygon area corresponding to \ ( ACDE \) area minus \ (\ triangle ABC \) area.
Code
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef double db;
const db eps = 1e-10;
const db pi = acos(-1.0);
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 100 + 10;
inline int dcmp(db x) {
if(fabs(x) < eps) return 0;
return x > 0? 1: -1;
}
class Point {
public:
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
void input() {
scanf("%lf%lf", &x, &y);
}
bool operator<(const Point &a) const {
return (!dcmp(x - a.x))? dcmp(y - a.y) < 0: x < a.x;
}
bool operator==(const Point &a) const {
return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
}
db dis2(const Point a) {
return pow(x - a.x, 2) + pow(y - a.y, 2);
}
db dis(const Point a) {
return sqrt(dis2(a));
}
db dis2() {
return x * x + y * y;
}
db dis() {
return sqrt(dis2());
}
Point operator+(const Point a) {
return Point(x + a.x, y + a.y);
}
Point operator-(const Point a) {
return Point(x - a.x, y - a.y);
}
Point operator*(double p) {
return Point(x * p, y * p);
}
Point operator/(double p) {
return Point(x / p, y / p);
}
db dot(const Point a) {
return x * a.x + y * a.y;
}
db cross(const Point a) {
return x * a.y - y * a.x;
}
};
Point p[maxn];
int n;
db area() {
if(n < 3) return 0.0;
db ans = 0.0;
for(int i = 2; i < n; ++i) {
ans += (p[i] - p[1]).cross(p[i + 1] - p[1]);
}
return ans * 0.5;
}
int main() {
while(~scanf("%d", &n) && n) {
db s = 0;
for(int i = 1; i <= n; ++i) {
p[i].input();
}
s = area();
printf("%.1lf\n", fabs(s));
}
return 0;
}