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  • 简单几何(线段相交)+模拟 POJ 3449 Geometric Shapes

    题目传送门

    题意:给了若干个图形,问每个图形与哪些图形相交

    分析:题目说白了就是处理出每个图形的线段,然后判断是否相交。但是读入输出巨恶心,就是个模拟题加上线段相交的判断,我第一次WA不知道输出要按字母序输出,第二次WA是因为忘记多边形的最后一条线段,还好找到了,没有坚持的话就不会AC了。

    /************************************************
    * Author        :Running_Time
    * Created Time  :2015/10/31 星期六 13:38:11
    * File Name     :POJ_3449.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 = 1e5 + 10;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1e9 + 7;
    const double EPS = 1e-10;
    const double PI = acos (-1.0);
    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;
    }
    struct Point    {       //点的定义
        double x, y;
        Point () {}
        Point (double x, double y) : x (x), y (y) {}
        Point operator + (const Point &r) const {       //向量加法
            return Point (x + r.x, y + r.y);
        }
        Point operator - (const Point &r) const {       //向量减法
            return Point (x - r.x, y - r.y);
        }
        Point operator * (double p) const {       //向量乘以标量
            return Point (x * p, y * p);
        }
        Point operator / (double p) const {       //向量除以标量
            return Point (x / p, y / p);
        }
        bool operator < (const Point &r) const {       //点的坐标排序
            return x < r.x || (x == r.x && y < r.y);
        }
        bool operator == (const Point &r) const {       //判断同一个点
            return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
        }
    };
    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 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;
    }
    double polar_angle(Vector A)  {     //向量极角
        return atan2 (A.y, A.x);
    }
    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));
    }
    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 line_line_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 point_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 point_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_line_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 can_seg_seg_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 can_line_seg_inter(Point a1, Point a2, Point b1, Point b2)    {        //判断直线与线段相交,两点式
        double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1);
        return dcmp (c1 * c2) <= 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_triangle(Point a, Point b, Point c) {       //三角形面积,叉积
        return fabs (cross (b - a, c - a)) / 2.0;
    }
    double area_poly(Point *p, int n)   {       //多边形面积,叉积
        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 Seg  {
        Point a, b;
        Seg ()  {}
        Seg (Point a, Point b) : a (a), b (b)   {}
    };
    struct  Pic {
        char ch;
        int cnt;
        vector<Seg> seg;
        vector<int> ans;
        bool operator < (const Pic &r) const    {
            return ch < r.ch;
        }
    }p[33], pp[33];
    
    bool cmp(int i, int j)  {
        return p[i].ch < p[j].ch;
    }
    
    void run(int tot)  {
        for (int i=0; i<tot; ++i)   {
            p[i].cnt = 0;
            for (int j=0; j<tot; ++j)   {
                if (i == j) continue;
                int sz1 = p[i].seg.size (), sz2 = p[j].seg.size ();
                bool flag = false;
                for (int ii=0; ii<sz1; ++ii)   {
                    for (int jj=0; jj<sz2; ++jj)   {
                        if (can_seg_seg_inter (p[i].seg[ii].a, p[i].seg[ii].b, p[j].seg[jj].a, p[j].seg[jj].b)) {
                            flag = true;    p[i].cnt++; p[i].ans.push_back (j);
                            break;
                        }
                    }
                    if (flag)   break;
                }
            }
            sort (p[i].ans.begin (), p[i].ans.end (), cmp);
        }
    }
    
    int main(void)    {
        //freopen ("POJ_3449.in", "r", stdin);
        //freopen ("POJ_3449.out", "w", stdout);
        char str[22];
        int x[22], y[22];
        int tot = 0;
        int x1, y1, x2, y2, x3, y3, x4, y4;
        Point p1, p2, p3, p4;
        while (scanf ("%s", &str) == 1) {
            if (strcmp (str, ".") == 0)  break;
            else if (strcmp (str, "-") == 0) {
                run (tot);
                for (int i=0; i<tot; ++i)   pp[i] = p[i];
                sort (p, p+tot);
                for (int i=0; i<tot; ++i)   {
                    if (p[i].cnt == 0)  {
                        printf ("%c has no intersections
    ", p[i].ch);
                    }
                    else    {
                        if (p[i].cnt == 1)  {
                            printf ("%c intersects with %c
    ", p[i].ch, pp[p[i].ans[0]].ch);
                        }
                        else if (p[i].cnt == 2) {
                            printf ("%c intersects with %c and %c
    ", p[i].ch, pp[p[i].ans[0]].ch, pp[p[i].ans[1]].ch);
                        }
                        else    {
                            printf ("%c intersects with %c", p[i].ch, pp[p[i].ans[0]].ch);
                            int sz = p[i].ans.size ();
                            for (int j=1; j<sz-1; ++j)    {
                                printf (", %c", pp[p[i].ans[j]].ch);
                            }
                            printf (", and %c
    ", pp[p[i].ans[sz-1]].ch);
                        }
                    }
                    p[i].ans.clear ();  p[i].seg.clear ();
                }
                puts ("");
                tot = 0;    continue;
            }
    
            p[tot].ch = str[0];
            scanf ("%s", &str);
            if (str[0] == 's')  {
                scanf (" (%d,%d) (%d,%d)", &x1, &y1, &x3, &y3);
                //cal x2 x4
                p1 = Point (x1, y1), p3 = Point (x3, y3), p2, p4;
                Vector V = p3 - p1;
                double len = length (V);
                p2 = p1 + rotate (V, PI / 4) / sqrt (2.0);
                p4 = p1 + rotate (V, -PI / 4) / sqrt (2.0);
                p[tot].seg.push_back (Seg (p1, p2));
                p[tot].seg.push_back (Seg (p2, p3));
                p[tot].seg.push_back (Seg (p3, p4));
                p[tot].seg.push_back (Seg (p4, p1));
            }
            else if (str[0] == 'l') {
                scanf (" (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2);
                p[tot].seg.push_back (Seg (Point (x1, y1), Point (x2, y2)));
            }
            else if (str[0] == 't') {
                scanf (" (%d,%d) (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2, &x3, &y3);
                p[tot].seg.push_back (Seg (Point (x1, y1), Point (x2, y2)));
                p[tot].seg.push_back (Seg (Point (x2, y2), Point (x3, y3)));
                p[tot].seg.push_back (Seg (Point (x3, y3), Point (x1, y1)));
            }
            else if (str[0] == 'p') {
                int n;
                scanf ("%d", &n);
                for (int i=0; i<n; ++i) {
                    scanf (" (%d,%d)", &x[i], &y[i]);
                }
                for (int i=0; i<n-1; ++i)   {
                    p[tot].seg.push_back (Seg (Point (x[i], y[i]), Point (x[i+1], y[i+1])));
                    p[tot].seg.push_back (Seg (Point (x[n-1], y[n-1]), Point (x[0], y[0])));        //忘记加了
                }
            }
            else if (str[0] == 'r') {
                scanf (" (%d,%d) (%d,%d) (%d,%d)", &x1, &y1, &x2, &y2, &x3, &y3);
                //cal x4
                p1 = Point (x1, y1), p3 = Point (x3, y3), p2 = Point (x2, y2), p4;
                p4 = p1 + (p3 - p2);
                p[tot].seg.push_back (Seg (p1, p2));
                p[tot].seg.push_back (Seg (p2, p3));
                p[tot].seg.push_back (Seg (p3, p4));
                p[tot].seg.push_back (Seg (p4, p1));
            }
            tot++;
        }
    
        //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
    ";
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4925663.html
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