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

时间:2015-10-31 16:58:54      阅读:205      评论:0      收藏:0      [点我收藏+]

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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\n", p[i].ch);
                }
                else    {
                    if (p[i].cnt == 1)  {
                        printf ("%c intersects with %c\n", p[i].ch, pp[p[i].ans[0]].ch);
                    }
                    else if (p[i].cnt == 2) {
                        printf ("%c intersects with %c and %c\n", 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\n", 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.\n";

    return 0;
}

  

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

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

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