• 简单几何(点与线段的位置) POJ 2318 TOYS && POJ 2398 Toy Storage


    题目传送门

    题意:POJ 2318 有一个长方形,用线段划分若干区域,给若干个点,问每个区域点的分布情况

    分析:点和线段的位置判断可以用叉积判断。给的线段是排好序的,但是点是无序的,所以可以用二分优化。用到了叉积

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/23 星期五 11:38:18
    * File Name     :POJ_2318.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 = 5e3 + 10;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1e9 + 7;
    const double EPS = 1e-10;
    struct Point    {       //点的定义
        double x, y;
        Point (double x=0, double y=0) : x (x), y (y) {}
    };
    typedef Point Vector;       //向量的定义
    Point read_point(void)   {      //点的读入
        double x, y;
        scanf ("%lf%lf", &x, &y);
        return Point (x, y);
    }
    double polar_angle(Vector A)  {     //向量极角
        return atan2 (A.y, A.x);
    }
    double dot(Vector A, Vector B)  {       //向量点积
        return A.x * B.x + A.y * B.y;
    }
    double cross(Vector A, Vector B)    {       //向量叉积
        return A.x * B.y - A.y * B.x;
    }
    int dcmp(double x)  {       //三态函数,减少精度问题
        if (fabs (x) < EPS) return 0;
        else    return x < 0 ? -1 : 1;
    }
    Vector operator + (Vector A, Vector B)  {       //向量加法
        return Vector (A.x + B.x, A.y + B.y);
    }
    Vector operator - (Vector A, Vector B)  {       //向量减法
        return Vector (A.x - B.x, A.y - B.y);
    }
    Vector operator * (Vector A, double p)  {       //向量乘以标量
        return Vector (A.x * p, A.y * p);
    }
    Vector operator / (Vector A, double p)  {       //向量除以标量
        return Vector (A.x / p, A.y / p);
    }
    bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    bool operator == (const Point &a, const Point &b)   {       //判断同一个点
        return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
    }
    double length(Vector A) {       //向量长度,点积
        return sqrt (dot (A, A));
    }
    double angle(Vector A, Vector B)    {       //向量转角,逆时针,点积
        return acos (dot (A, B) / length (A) / length (B));
    }
    double area_triangle(Point a, Point b, Point c) {       //三角形面积,叉积
        return fabs (cross (b - a, c - a)) / 2.0;
    }
    Vector rotate(Vector A, double rad) {       //向量旋转,逆时针
        return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
    }
    Vector nomal(Vector A)  {       //向量的单位法向量
        double len = length (A);
        return Vector (-A.y / len, A.x / len);
    }
    Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点,参数方程
        Vector U = p - q;
        double t = cross (W, U) / cross (V, W);
        return p + V * t;
    }
    double dis_to_line(Point p, Point a, Point b)   {       //点到直线的距离,两点式
        Vector V1 = b - a, V2 = p - a;
        return fabs (cross (V1, V2)) / length (V1);
    }
    double dis_to_seg(Point p, Point a, Point b)    {       //点到线段的距离,两点式
      
        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);
    }
    Point point_proj(Point p, Point a, Point b)   {     //点在直线上的投影,两点式
        Vector V = b - a;
        return a + V * (dot (V, p - a) / dot (V, V));
    }
    bool inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交,两点式
        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)    {       //判断点在线段上,两点式
        return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
    }
    double area_poly(Point *p, int n)   {       //多边形面积
        double ret = 0;
        for (int i=1; i<n-1; ++i)   {
            ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
        }
        return ret / 2;
    }
    struct Line {
        Point s, e;
        Line () {}
        Line (Point s, Point e) : s (s), e (e) {}
    };
    Line board[N];
    Point toy[N];
    Point b, e;
    int ans[N];
    
    int main(void)    {
        int n, m;
        bool fir = true;
        while(scanf ("%d", &n) == 1)    {
            if (!n) break;
            if (fir)    fir = false;
            else    puts ("");
            scanf ("%d", &m);
            b = read_point ();
            e = read_point ();
            double U, L;
            for (int i=0; i<n; ++i) {
                scanf ("%lf%lf", &U, &L);
                board[i] = Line (Point (U, b.y), Point (L, e.y));
            }
            board[n] = Line (Point (e.x, b.y), Point (e.x, e.y));
            for (int i=0; i<m; ++i) {
                toy[i] = read_point ();
            }
            memset (ans, 0, sizeof (ans));
            for (int i=0; i<m; ++i) {
                if (toy[i].x < b.x || toy[i].x > e.x || toy[i].y > b.y || toy[i].y < e.y)   continue;
                int l = 0, r = n, mid, tmp = 0;
                while (l <= r)   {
                    mid = (l + r) >> 1;
                    if (cross (board[mid].s - toy[i], board[mid].e - toy[i]) < 0)   {
                        tmp = mid;
                        r = mid - 1;
                    }
                    else    l = mid + 1;
                }
                ans[tmp]++;
            }
            for (int i=0; i<=n; ++i)    {
                printf ("%d: %d
    ", i, ans[i]);
            }
        }
    
        return 0;
    }
    

    题目传送门

    题意:POJ 2398 和上面那题没什么区别,回答的问题不同,线段无序先排序。

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/23 星期五 11:38:18
    * File Name     :POJ_2398.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 = 1e3 + 10;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1e9 + 7;
    const double EPS = 1e-10;
    struct Point    {       //点的定义
        double x, y;
        Point (double x=0, double y=0) : x (x), y (y) {}
    };
    typedef Point Vector;       //向量的定义
    Point read_point(void)   {      //点的读入
        double x, y;
        scanf ("%lf%lf", &x, &y);
        return Point (x, y);
    }
    double polar_angle(Vector A)  {     //向量极角
        return atan2 (A.y, A.x);
    }
    double dot(Vector A, Vector B)  {       //向量点积
        return A.x * B.x + A.y * B.y;
    }
    double cross(Vector A, Vector B)    {       //向量叉积
        return A.x * B.y - A.y * B.x;
    }
    int dcmp(double x)  {       //三态函数,减少精度问题
        if (fabs (x) < EPS) return 0;
        else    return x < 0 ? -1 : 1;
    }
    Vector operator + (Vector A, Vector B)  {       //向量加法
        return Vector (A.x + B.x, A.y + B.y);
    }
    Vector operator - (Vector A, Vector B)  {       //向量减法
        return Vector (A.x - B.x, A.y - B.y);
    }
    Vector operator * (Vector A, double p)  {       //向量乘以标量
        return Vector (A.x * p, A.y * p);
    }
    Vector operator / (Vector A, double p)  {       //向量除以标量
        return Vector (A.x / p, A.y / p);
    }
    bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    bool operator == (const Point &a, const Point &b)   {       //判断同一个点
        return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
    }
    double length(Vector A) {       //向量长度,点积
        return sqrt (dot (A, A));
    }
    double angle(Vector A, Vector B)    {       //向量转角,逆时针,点积
        return acos (dot (A, B) / length (A) / length (B));
    }
    double area_triangle(Point a, Point b, Point c) {       //三角形面积,叉积
        return fabs (cross (b - a, c - a)) / 2.0;
    }
    Vector rotate(Vector A, double rad) {       //向量旋转,逆时针
        return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
    }
    Vector nomal(Vector A)  {       //向量的单位法向量
        double len = length (A);
        return Vector (-A.y / len, A.x / len);
    }
    Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点,参数方程
        Vector U = p - q;
        double t = cross (W, U) / cross (V, W);
        return p + V * t;
    }
    double dis_to_line(Point p, Point a, Point b)   {       //点到直线的距离,两点式
        Vector V1 = b - a, V2 = p - a;
        return fabs (cross (V1, V2)) / length (V1);
    }
    double dis_to_seg(Point p, Point a, Point b)    {       //点到线段的距离,两点式
      
        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);
    }
    Point point_proj(Point p, Point a, Point b)   {     //点在直线上的投影,两点式
        Vector V = b - a;
        return a + V * (dot (V, p - a) / dot (V, V));
    }
    bool inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交,两点式
        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)    {       //判断点在线段上,两点式
        return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
    }
    double area_poly(Point *p, int n)   {       //多边形面积
        double ret = 0;
        for (int i=1; i<n-1; ++i)   {
            ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
        }
        return ret / 2;
    }
    struct Line {
        Point s, e;
        Line () {}
        Line (Point s, Point e) : s (s), e (e) {}
    };
    bool cmp(Line A, Line B)    {
        return A.s.x < B.s.x || (A.s.x == B.s.x && A.e.x < B.e.x);
    }
    Line board[N];
    Point toy[N];
    Point b, e;
    int ans[N];
    
    int main(void)    {
        int n, m;
        while(scanf ("%d", &n) == 1)    {
            if (!n) break;
            scanf ("%d", &m);
            b = read_point ();
            e = read_point ();
            double U, L;
            for (int i=0; i<n; ++i) {
                scanf ("%lf%lf", &U, &L);
                board[i] = Line (Point (U, b.y), Point (L, e.y));
            }
            board[n] = Line (Point (e.x, b.y), Point (e.x, e.y));
            sort (board, board+1+n, cmp);
            for (int i=0; i<m; ++i) {
                toy[i] = read_point ();
            }
            memset (ans, 0, sizeof (ans));
            for (int i=0; i<m; ++i) {
                if (toy[i].x < b.x || toy[i].x > e.x || toy[i].y > b.y || toy[i].y < e.y)   continue;
                int l = 0, r = n, mid, tmp = 0;
                while (l <= r)   {
                    mid = (l + r) >> 1;
                    if (cross (board[mid].s - toy[i], board[mid].e - toy[i]) < 0)   {
                        tmp = mid;
                        r = mid - 1;
                    }
                    else    l = mid + 1;
                }
                ans[tmp]++;
            }
            puts ("Box");
            for (int i=1; i<=m; ++i)    {
                int cnt = 0;
                for (int j=0; j<=n; ++j)    {
                    if (ans[j] == i)    ++cnt;
                }
                if (cnt)    {
                    printf ("%d: %d
    ", i, cnt);
                }
            }
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4905285.html
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