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EOJ 1058. 挤模具 (多边形面积)

题目链接:1058. 挤模具

题意

给出模具的底和体积,求模具的高。

思路

模具的底为多边形,因此求出多边形面积,用体积除以底的面积就是答案。

多边形的面积求解见 EOJ 1127. 多边形面积(计算几何)

代码

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#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;
db v;
for(int i = 1; i <= n; ++i) {
p[i].input();
}
scanf("%lf", &v);
s = area();
printf("BAR LENGTH: %.2lf\n", v / fabs(s));
}
return 0;
}

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