hdu 1665 That Nice Euler Circuit(欧拉定理)

输入n个点，然后从第一个点开始，依次链接点i->点i+1，最后回到第一点（输入中的点n），求得到的图形将平面分成了多少部分。
根据欧拉定理 v_num + f_num - e_num = 2可知，求出点数跟边数便能求出平面数。
#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<fstream>

#include<sstream>

#include<bitset>

#include<vector>

#include<string>

#include<cstdio>

#include<cmath>

#include<stack>

#include<queue>

#include<stack>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define debug puts("**debug**")

#define LL long long

#define PB push_back

#define eps 1e-10

using namespace std;

struct Point

{

    double x, y;

    Point (double x=0, double y=0):x(x), y(y) {}

};

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }

Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b)

{

    return a.x < b.x || (a.x == b.x && a.y < b.y);

}

int dcmp(double x)

{

    if(fabs(x) < eps) return 0;

    return x < 0 ? -1 : 1;

}

bool operator == (const Point& a, const Point& b)

{

    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;

}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }

double Length(Vector A) { return sqrt(Dot(A, A)); }

double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

double Area2(Vector A, Vector B, Vector C) { return Cross(B-A, C-A); }

//向量逆时针旋转

Vector Rotate(Vector A, double rad)

{

    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));

}

//求直线p+tv和q+tw的交点 Cross(v, w) == 0无交点

Point GetLineIntersection(Point p, Vector v, Point q, Vector w)

{

    Vector u = p-q;

    double t = Cross(w, u) / Cross(v, w);

    return p + v*t;

}

//点p到直线ab的距离

double DistanceToLine(Point p, Point a, Point b)

{

    Vector v1 = b - a, v2 = p - a;

    return fabs(Cross(v1, v2)) / Length(v1);//如果不带fabs 得到的是有向距离

}

//点p到线段ab的距离

double DistanceToSegment(Point p, Point a, Point b)

{

    if(a == b) return Length(p-a);

    Vector v1 = b-a, v2 = p-a, v3 = p-b;

    if(dcmp(Dot(v1, v2) < 0)) return Length(v2);

    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);

    else return fabs(Cross(v1, v2)) / Length(v1);

}

//点p在直线ab上的投影

Point GetLineProjection(Point p, Point a, Point b)

{

    Vector v = b-a;

    return a + v*(Dot(v, p-a) / Dot(v, v));

}

//点段相交判定

bool SegmentItersection(Point a1, Point a2, Point b1, Point b2)

{

    double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1),

    c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);

    return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;

}

//点在线段上

bool OnSegment(Point p, Point a1, Point a2)

{

    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;

}

//多变形面积

double PolygonArea(Point* p, int n)

{

    double ret = 0;

    FF(i, 1, n) ret += Cross(p[i]-p[0], p[i+1]-p[0]);

    return ret/2;

}

Point read_point()

{

    Point a;

    scanf("%lf%lf", &a.x, &a.y);

    return a;

}

const int maxn = 301;

Point p[maxn], v[maxn*1000];

int main()

{

    int n, kase = 1;

    while(scanf("%d", &n), n)

    {

        REP(i, n)

        {

            scanf("%lf%lf", &p[i].x, &p[i].y);

            v[i] = p[i];

        }

        int cnt = n-1, e_num = n-1;//原有边数

        //枚举所有边，求出相交出来的新的点

        REP(i, n-1) FF(j, i+1, n-1)

        if(SegmentItersection(p[i], p[i+1], p[j], p[j+1]))

        v[cnt++] = GetLineIntersection(p[i], p[i+1]-p[i], p[j], p[j+1]-p[j]);

        sort(v, v+cnt);

        int v_num = unique(v, v+cnt) - v;//去重点

        REP(i, v_num) REP(j, n-1)

        if(OnSegment(v[i], p[j], p[j+1])) e_num++;//线段被切割

        printf("Case %d: There are %d pieces.\n", kase++, e_num + 2 - v_num);

    }

    return 0;

}
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hdu 1665 That Nice Euler Circuit(欧拉定理)

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