• 简单几何(求划分区域) LA 3263 That Nice Euler Circuit


    题目传送门

    题意:一笔画,问该图形将平面分成多少个区域

    分析:训练指南P260,欧拉定理:平面图定点数V,边数E,面数F,则V + F - E =  2。那么找出新增的点和边就可以了。用到了判断线段相交,求交点,判断点在线上

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/22 星期四 09:10:09
    * File Name     :LA_3263.cpp
     ************************************************/
    
    #include <cstdio>
    #include <algorithm>
    #include <iostream>
    #include <sstream>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <vector>
    #include <queue>
    #include <deque>
    #include <stack>
    #include <list>
    #include <map>
    #include <set>
    #include <bitset>
    #include <cstdlib>
    #include <ctime>
    using namespace std;
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    typedef long long ll;
    const int N = 300 + 10;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1e9 + 7;
    const double EPS = 1e-10;
    struct Point    {
        double x, y;
        Point (double x=0, double y=0) : x (x), y (y) {}
    };
    typedef Point Vector;
    double dot(Vector A, Vector B)  {
        return A.x * B.x + A.y * B.y;
    }
    double cross(Vector A, Vector B)    {
        return A.x * B.y - A.y * B.x;
    }
    int dcmp(double x)  {
        if (fabs (x) < EPS) return 0;
        else    return x < 0 ? -1 : 1;
    }
    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);
    }
    bool operator == (const Point &a, const Point &b)   {
        return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
    }
    double length(Vector A) {
        return sqrt (dot (A, A));
    }
    Point point_inter(Point p, Vector V, Point q, Vector W)    {
        Vector U = p - q;
        double t = cross (W, U) / cross (V, W);
        return p + V * t;
    }
    Point point_proj(Point p, Point a, Point b)   {
        Vector V = b - a;
        return a + V * (dot (V, p - a) / dot (V, V));
    }
    bool inter(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;
    }
    bool on_seg(Point p, Point a1, Point a2)    {
        return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
    }
    Point P[N], V[N*N];
    
    int main(void)    {
        int n, cas = 0;
        while (scanf ("%d", &n) == 1)  {
            if (!n) break;
            for (int i=0; i<n; ++i) {
                scanf ("%lf%lf", &P[i].x, &P[i].y);
                V[i] = P[i];
            }
            n--;
            int v = n, e = n;
            for (int i=0; i<n; ++i) {
                for (int j=i+1; j<n; ++j)   {
                    if (inter (P[i], P[i+1], P[j], P[j+1])) {
                        V[v++] = point_inter (P[i], P[i+1] - P[i], P[j], P[j+1] - P[j]);
                    }
                }
            }
            sort (V, V+v);
            v = unique (V, V+v) - V;
            for (int i=0; i<v; ++i) {
                for (int j=0; j<n; ++j) {
                    if (on_seg (V[i], P[j], P[j+1]))    e++;
                }
            }
            printf ("Case %d: There are %d pieces.
    ", ++cas, e + 2 - v);
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4900270.html
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