sgu209：Areas(计算几何)

意甲冠军：

给一些直。这架飞机被分成了很多这些线性块。每个块的需求面积封闭曲线图。

分析：

①我们应要求交点22的直线；

②每行上的交点的重排序，借此来离散一整行（正反两条边）；

③对于连向一个点的几条线段，对它们进行极角排序，相邻的两条线段我们给它们之间连一条边，我们脑补一下应该能够知道如何能够保证逆时针连边；

④找循环，利用叉积求面积。

ps.ps. vectorvector的调试真心不爽…

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#define pb push_back
#include <utility>
#define fi first
#define se second
#define mp make_pair
#include <stack>
using namespace std;
const int MAXN = 100;
const int MAXE = MAXN*MAXN<<1;
const double eps = 1e-8;
typedef double DB;

int n;
#define sqr(x) ((x)*(x))
struct pot
{
double x, y;
pot(double x = 0, double y = 0):x(x), y(y){}
inline double len() {return sqrt(sqr(x)+sqr(y));}
inline void read() {scanf("%lf%lf", &x, &y);}
inline pot unit()
{
double d = len();
return pot(x/d, y/d);
}
inline double ang() {return atan2(y, x);}
};
typedef pot vec;
struct Line
{
pot p;
vec v;
Line(){}
Line(pot p, vec v):p(p), v(v){}
}lines[MAXN];
vector<pot> points;

struct Edge
{
int sz;
int head[MAXE], to[MAXE], ne[MAXE];
Edge()
{
sz = 0;
memset(head, -1, sizeof(head));
}
inline void add(int u, int v)
{
to[sz] = v;ne[sz] = head[u];
head[u] = sz++;
}
}E;

int next[MAXE];

bool vis[MAXE];

vector<DB> ans;

inline int dcmp(double x)
{
if(fabs(x) <= eps) return 0;
else return x>0?1:-1;
}

inline pot operator + (const vec &a, const vec &b) {return vec(a.x+b.x, a.y+b.y);}
inline pot operator - (const vec &a, const vec &b) {return vec(a.x-b.x, a.y-b.y);}
inline double operator * (const vec &a, const vec &b) {return a.x*b.x+a.y*b.y;}
inline double operator ^ (const vec &a, const vec &b) {return a.x*b.y-a.y*b.x;}
inline pot operator * (const vec &a, double k) {return vec(a.x*k, a.y*k);}
inline pot operator / (const vec &a, double k) {return vec(a.x/k, a.y/k);}
inline bool operator < (const vec &a, const vec &b) {return dcmp(a.x-b.x) < 0 || (dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) < 0);}
inline bool operator == (const vec &a, const vec &b) {return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}

inline pot get_line_intersection(const pot &p, const vec &v, const pot &q, const vec &w)
{
vec u = p-q;
double t = (w^u)/(v^w);
return p+v*t;
}

inline bool parallel(const Line &a, const Line &b) {return dcmp(a.v^b.v) == 0;}

inline int get_pot_id(const pot &p) {return lower_bound(points.begin(), points.end(), p)-points.begin();}

inline bool cmp(DB a, DB b) {return dcmp(a-b) == 0;}

inline void init()
{
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
pot a, b;
a.read();b.read();
lines[i] = Line(a, b-a);
}
for(int i = 0; i < n; ++i)
for(int j = i+1; j < n; ++j)
if(!parallel(lines[i], lines[j]))
points.pb(get_line_intersection(lines[i].p, lines[i].v, lines[j].p, lines[j].v));
sort(points.begin(), points.end());
points.erase(unique(points.begin(), points.end()), points.end());
for(int i = 0; i < n; ++i)
{
vector<DB> nodes;
vec d = lines[i].v.unit();
for(int j = 0; j < n; ++j)
if(!parallel(lines[i], lines[j]))
{
pot inter = get_line_intersection(lines[i].p, lines[i].v, lines[j].p, lines[j].v);
nodes.pb((inter-lines[i].p)*d);
}
sort(nodes.begin(), nodes.end());
nodes.erase(unique(nodes.begin(), nodes.end(), cmp), nodes.end());
for(int j = 1, sz = nodes.size(); j < sz; ++j)
{
int a = get_pot_id(lines[i].p+d*nodes[j]);
int b = get_pot_id(lines[i].p+d*nodes[j-1]);
E.add(a, b);E.add(b, a);
}
}
memset(next, -1, sizeof(next));
for(int i = 0, sz = points.size(); i < sz; ++i)
{
vector<pair<DB, int> > PA;
for(int j = E.head[i]; j != -1; j = E.ne[j])
PA.pb(mp((points[E.to[j]]-points[i]).ang(), j));
sort(PA.begin(), PA.end());
for(int j = 0, sz = PA.size(); j < sz; ++j)
next[PA[(j+1)%sz].se^1] = PA[j].se;
}
}

inline void work()
{
for(int i = 0; i < E.sz; ++i)
if(!vis[i])
{
stack<int> s;
int j = i;
do
{
if(!s.empty() && (s.top()^1) == j)
s.pop();
else s.push(j);
vis[j] = true;
j = next[j];
if(j == -1) break;
}while(!vis[j]);
if(i == j)
{
DB area = 0;
while(!s.empty())
{
area += (points[E.to[s.top()^1]]^points[E.to[s.top()]]);
s.pop();
}
area *= 0.5;
if(dcmp(area) > 0)
ans.pb(area);
}
}
}

inline void print()
{
printf("%d\n", ans.size());
sort(ans.begin(), ans.end());
for(int i = 0, sz = ans.size(); i < sz; ++i)
printf("%.4lf\n", ans[i]);
}

int main()
{
init();
work();
print();
return 0;
}