POJ 1066.Treasure Hunt

537


题目



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Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-art technology they are able to determine that the lower floor of the pyramid is constructed from a series of straightline walls, which intersect to form numerous enclosed chambers. Currently, no doors exist to allow access to any chamber. This state-of-the-art technology has also pinpointed the location of the treasure room. What these dedicated (and greedy) archeologists want to do is blast doors through the walls to get to the treasure room. However, to minimize the damage to the artwork in the intervening chambers (and stay under their government grant for dynamite) they want to blast through the minimum number of doors. For structural integrity purposes, doors should only be blasted at the midpoint of the wall of the room being entered. You are to write a program which determines this minimum number of doors.
An example is shown below:



The input will consist of one case. The first line will be an integer n (0 <= n <= 30) specifying number of interior walls, followed by n lines containing integer endpoints of each wall x1 y1 x2 y2 . The 4 enclosing walls of the pyramid have fixed endpoints at (0,0); (0,100); (100,100) and (100,0) and are not included in the list of walls. The interior walls always span from one exterior wall to another exterior wall and are arranged such that no more than two walls intersect at any point. You may assume that no two given walls coincide. After the listing of the interior walls there will be one final line containing the floating point coordinates of the treasure in the treasure room (guaranteed not to lie on a wall).


Print a single line listing the minimum number of doors which need to be created, in the format shown below.


7
20 0 37 100
40 0 76 100
85 0 0 75
100 90 0 90
0 71 100 61
0 14 100 38
100 47 47 100
54.5 55.4


Number of doors = 2




题解



一开始没仔细读题,没有考虑中点,直接求了向各个方向相交最小值
(一般按照套路也只能这样算)

注意到中点后,推了一下,发现好像不需要刻意考虑中点
大概理由如下:
从任意一个位置到宝藏处,要通过的墙是确定的,不需要可以考虑墙的先后顺序
如果直接走过来不穿这个墙,就说明这个墙对我们毫无影响
如果直接走过来穿了这个墙,不管先穿后穿只打一个洞就行了(相当于通过了这一层)

因此直接跑就行了

最后记得记上边界墙
同时判断相交不考虑端点情况
特判 n==0

代码


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#include <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
using namespace std;

#define Log(format, ...) // printf(format, ##__VA_ARGS__)

/* 计算几何模板 */
const double eps = 1e-8;
const double INF = 9e50;
typedef complex<double> Vector;
typedef Vector Point;
struct Segment {
    Point a, b;
    Segment() {}
    Segment(Point _a, Point _b) : a(_a), b(_b) {}
    Vector toVector() { return b - a; }
};

inline Point read_Point() {
    double x, y;
    scanf("%lf%lf", &x, &y);
    return Point(x, y);
}
inline int sgn(const double &x) {
    if (fabs(x) < eps)
        return 0;
    return x > 0 ? 1 : -1;
}
inline bool operator==(const Vector &vec1, const Vector &vec2) {
    return sgn(vec1.real() - vec2.real()) == 0 &&
           sgn(vec1.imag() - vec2.imag()) == 0;
}
inline bool operator!=(const Vector &vec1, const Vector &vec2) {
    return !(vec1 == vec2);
}

inline int Cross(const Vector &a, const Vector &b) {
    return a.real() * b.imag() - a.imag() * b.real();
}
inline int Dot(const Vector &a, const Vector &b) {
    return a.real() * b.real() + a.imag() * b.imag();
}

inline double Distance(const Point &a, const Point &b) { return abs(a - b); }

inline int Point_Segment(const Vector &p, const Segment &L) {
    return sgn(Cross(L.b - L.a, p - L.a));
}
inline int Segment_Segment(const Segment L1, const Segment L2) {
    double a = L1.b.real() - L1.a.real();
    double b = L2.b.real() - L2.a.real();
    double c = L1.b.imag() - L1.a.imag();
    double d = L2.b.imag() - L2.a.imag();
    double f = a * d - b * c;

    // 平行或重叠
    if (sgn(f) == 0)
        return -1;

    double g = L2.b.real() - L1.a.real();
    double h = L2.b.imag() - L1.a.imag();
    double t = (d * g - b * h) / f;
    double s = (-c * g + a * h) / f;

    // 在延长线上
    if (t <= 0 || t >= 1 || s <= 0 || s >= 1)
        return 0;

    // 线段相交
    return 1;
}

const int maxn = 35;
int n;
Segment segments[maxn];

int calc(Segment L) {
    int cnt = 0;
    for (int i = 0; i < n; ++i)
        cnt += (Segment_Segment(L, segments[i]) > 0);
    return cnt;
}

int main() {
    while (~scanf("%d", &n)) {
        for (int i = 0; i < n; ++i) {
            segments[i].a = read_Point();
            segments[i].b = read_Point();
        }
        Point target = read_Point();

        int ans = 9999999;
        for (int i = 0; i < n; i++) {
            ans = min(ans, calc(Segment(target, segments[i].a)));
            ans = min(ans, calc(Segment(target, segments[i].b)));
        }
        if (n == 0)
            ans = 0;
        printf("Number of doors = %d\n", ans + 1);
    }
    return 0;
}
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