[CQOI2005]三角形面积并(计算几何+扫描线)

题目描述

题解

这题思维难度不高，如果能想到扫描线应该很快就能想到正解，就是代码比较毒瘤。

首先在每个三角形的顶点和没两个三角形的交点建一条平行于y轴的扫描线，然后图形就被分成了若干的梯形（三角形），挨个求这些梯形的面积即可。

设一个梯形的上底是a，下底是b，高是h，则$S=frac{(a+b) imes h}{2}=中位线长度 imes h$，我们只需求出每个梯形的中位线长度即可。

对于两条相邻的扫描线x1,x2，它们之间的梯形的中位线就是$frac{x1+x2}{2}$截所有三角形所得的距离，因为数据小所以我们可以不用线段树维护。

具体实现细节建议用向量而非解析式。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const long double eps = 1e-8;
const long double inf = 2000000;
struct Point{
    long double x, y;
    Point operator + (Point v) {
        Point ret = Point{x + v.x, y + v.y};
        return ret;
    }
    Point operator - (Point v) {
        Point ret = Point{x - v.x, y - v.y};
        return ret;
    }
    Point operator * (long double v) {
        Point ret = Point{x * v, y * v};
        return ret;
    }
}t[110][4];
long double p[100000];
int tot;
int n;
inline int sgn(long double x) {
    if (fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}
inline long double dot(Point a, Point b) {
    return a.x * b.x + a.y * b.y;
}
inline long double cross(Point a, Point b) {
    return a.x * b.y - a.y * b.x;
}
inline bool is_intersect(Point a, Point b, Point c, Point d) {//判断是否相交 
    long double f1 = cross(b - a, c - a);
    long double f2 = cross(b - a, d - a);
    long double f3 = cross(d - c, a - c);
    long double f4 = cross(d - c, b - c);
    return (f1 * f2 < 0 && f3 * f4 < 0);
}
inline Point get_intersect(Point a, Point b, Point c, Point d) {
    long double s1 = cross(c - a, d - c), s2 = cross(b - a, d - c);
    return a + (b - a) * (s1 / s2);
}
bool cmp(Point x, Point y) {
    return x.x < y.x;
}
long double ans;
Point seg[10010];
inline long double solve(long double x) {
    Point P = Point{x, -inf}, Q = Point{x, inf};
    int m = 0;
    for (int i = 1; i <= n; i++) {
        int top = 0;
        long double y[3];
        for (int j = 1; j <= 3; j++) {
            int nxt_j = j + 1;
            if (nxt_j == 4) nxt_j = 1;
            if (is_intersect(t[i][j], t[i][nxt_j], P, Q)) {
                y[++top] = get_intersect(t[i][j], t[i][nxt_j], P, Q).y;
            }
        }
        if (top == 2) seg[++m] = Point{min(y[1], y[2]), max(y[1], y[2])};
    }
    sort(seg + 1, seg + m + 1, cmp);
    long double l = -inf, r = -inf, ret = 0;
    for (int i = 1; i <= m; i++) {
        if (sgn(seg[i].x - r) > 0) ret += r - l, l = seg[i].x;
        r = max(r, seg[i].y);
    }
    ret += r - l;
    return ret;
}
int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= 3; ++j) {
            cin >> t[i][j].x >> t[i][j].y;
            p[++tot] = t[i][j].x;
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            //枚举两个三角形求它们边的交点 
            for (int k = 1; k <= 3; ++k) {
                for (int l = 1; l <= 3; l++) {
                    int nxt_k = k + 1;
                    if (nxt_k == 4) nxt_k = 1;
                    int nxt_l = l + 1;
                    if (nxt_l == 4) nxt_l = 1;
                    if (is_intersect(t[i][k], t[i][nxt_k], t[j][l], t[j][nxt_l])) {
                        p[++tot] = get_intersect(t[i][k], t[i][nxt_k], t[j][l], t[j][nxt_l]).x;
                    }
                }
            }
        }
    }
    sort(p + 1, p + 1 + tot);
    for (int i = 2; i <= tot; i++) {
        if (sgn(p[i] - p[i - 1]) != 0) {
            ans += (p[i] - p[i - 1]) * solve((p[i] + p[i - 1]) / 2);//算梯形面积 
        }
    }
    printf("%.2Lf", ans-eps);//卡精度
    return 0;
}