[hdu4629 Burning]三角形面积并，扫描线

题意：给n个三角形，分别求覆盖1次~n次的总面积

思路： 对每个y坐标作一条平行于x轴的直线，按直线从下往上处理，每两条直线之间为若干梯形（也可以是三角形）首尾相连的情况，从左扫到右时，用一个变量cnt记录当前区域被覆盖的次数，遇到入边cnt++，遇到出边cnt--，边扫边更新答案。入边表示这条边的右边在三角形内部，出边表示这条边的右边在三角形的外部。思路并不复杂，只是代码稍长点。

 

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

const int maxn = 57;

int dcmp(double a) {
    if (abs(a) < EPS) return 0;
    return a > 0? 1 : -1;
}

struct Point {
    double x, y;
    void read() {
        int x, y;
        scanf("%d%d", &x, &y);
        this->x = x;
        this->y = y;
    }
    Point(double x, double y) {
        this->x = x;
        this->y = y;
    }
    Point() {}
    Point operator - (const Point &that) const {
        return Point(this->x - that.x, this->y - that.y);
    }
    bool operator < (const Point &that) const {
        return dcmp(this->y - that.y) < 0 || dcmp(this->y - that.y) == 0 &&
                dcmp(this->x - that.x) < 0;
    }
    Point operator * (const double &m) const {
        return Point(this->x * m, this->y * m);
    }
};
struct Segment {
    Point a, b;
    Segment(Point a, Point b) {
        this->a = a;
        this->b = b;
    }
    Segment() {}
    bool operator < (const Segment &that) const {
        return dcmp(this->a.x + this->b.x - that.a.x - that.b.x) < 0;
    }
};
Point point[maxn * maxn * 9];
Segment seg[maxn * 3];
int type[maxn * 3];
double ans[maxn];

double cross(const Point &a, const Point &b) {
    return a.x * b.y - a.y * b.x;
}

bool onMid(Point a, Point b, Point c) {
    return dcmp(cross(c - a, b - a)) == 0;
}
bool onLeft(Point a, Point b, Point c) {
    return dcmp(cross(c - a, b - a)) < 0;
}
bool Intersect(Segment A, Segment B) {
    int r1 = dcmp(cross(A.b - A.a, B.a - A.a));
    int r2 = dcmp(cross(A.b - A.a, B.b - A.a));
    int r3 = dcmp(cross(B.b - B.a, A.a - B.a));
    int r4 = dcmp(cross(B.b - B.a, A.b - B.a));
    return (r1 ^ r2) || (r3 ^ r4);
}
Point getLineIntersection(Segment A, Segment B) {
    Point u = A.a - B.a, v = A.b - A.a, w = B.b - B.a;
    double t = cross(w, u) / cross(v, w);
    return A.a - v * -t;
}
double Area(Segment A, Segment B) {
    return fabs((A.a.x - B.a.x + A.b.x - B.b.x) * (A.a.y - A.b.y) / 2);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, n, cs, cp;
    cin >> T;
    while (T --) {
        cin >> n;
        cp = cs = 0;
        for (int i = 0; i < n; i ++) {
            Point p[3];
            for (int j = 0; j < 3; j ++) p[j].read();
            if (onMid(p[0], p[1], p[2])) continue;
            sort(p, p + 3);
            seg[cs ++] = Segment(p[0], p[1]);
            seg[cs ++] = Segment(p[0], p[2]);
            seg[cs ++] = Segment(p[1], p[2]);
            if (onLeft(p[0], p[2], p[1])) {
                type[cs - 3] = type[cs - 1] = 1;
                type[cs - 2] = -1;
            }
            else {
                type[cs - 3] = type[cs - 1] = -1;
                type[cs - 2] = 1;
            }
        }
        for (int i = 0; i < cs; i ++) {
            for (int j = i + 1; j < cs; j ++) {
                if (Intersect(seg[i], seg[j]))
                    point[cp ++] = getLineIntersection(seg[i], seg[j]);
            }
        }
        sort(point, point + cp);
        vector<double> Y;
        for (int i = 0; i < cp; i ++) {
            if (!i || dcmp(point[i].y - point[i - 1].y) != 0) Y.pb(point[i].y);
        }
        fillchar(ans, 0);
        for (int i = 1; i < Y.size(); i ++) {
            vector<pair<Segment, int> > S;
            for (int j = 0; j < cs; j ++) {
                if (dcmp(seg[j].a.y - Y[i - 1]) <= 0 && dcmp(seg[j].b.y - Y[i]) >= 0) {
                    Point a = seg[j].a, b = seg[j].b;
                    double d = (b.x - a.x) / (b.y - a.y);
                    double x1 = a.x + d * (Y[i - 1] - a.y), x2 = a.x + d * (Y[i] - a.y);
                    S.pb(mp(Segment(Point(x1, Y[i - 1]), Point(x2, Y[i])), type[j]));
                }
            }
            sort(all(S));
            int cnt = 0;
            for (int j = 0; j < S.size(); j ++) {
                if (cnt) ans[cnt] += Area(S[j - 1].X, S[j].X);
                cnt += S[j].Y;
            }
        }
        for (int i = 1; i <= n; i ++) {
            printf("%.10f\n", ans[i]);
        }
    }

    return 0;
}