POJ1066 Treasure Hunt（线段相交）

Posted 2020-09-22 hyp1231

题目链接：

　　http://poj.org/problem?id=1066

题目描述：

Treasure Hunt

 

Description

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: 

Input

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).

Output

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

Sample Input

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 

Sample Output

Number of doors = 2 

题目大意：

　　给定一些在 (0,0) 和 (100,100) 之间的线段，在房间的墙的中点可以开门，问最少开多少门，可以从边上到达目标点

思路：

　　枚举边上的整点（但不能是给出的线段的端点）和整点之间带0.5的点作为起点，与目标点连成线段，与多少线段相交即需要开多少门，因为若与这些线段相交，则一定要在这些线段的某个墙的中点开门，两者等价

 

代码：

  1 #include <cmath>
  2 #include <algorithm>
  3 #include <cstdio>
  4 #include <iostream>
  5 #include <cstring>
  6 using namespace std;
  7 
  8 const double EPS = 1e-10;    //精度系数
  9 const int N = 32;
 10 const int M = 101;
 11 const int INF = 0x3f3f3f3f;
 12 
 13 struct Point {
 14     double x, y;
 15     Point(double x = 0, double y = 0) :x(x), y(y) {}
 16 };    //点的定义
 17 
 18 typedef Point Vector;    //向量的定义
 19 
 20 Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }    //向量减法
 21 
 22 int dcmp(double x) {
 23     if (fabs(x) < EPS)return 0; else return x < 0 ? -1 : 1;
 24 }    //与0的关系
 25 
 26 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }    //向量叉乘
 27 
 28 bool SegmentProperIntersection(Point& a1, Point& a2, Point& b1, Point& b2) {
 29     double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
 30         c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
 31     return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;
 32 }    //判断两线段相交（恰有一个公共点且不在端点）
 33 
 34 Point A[N], B[N], T;
 35 bool edge[M][M];
 36 
 37 int main() {
 38     int n, a1, b1, a2, b2, ans = INF;
 39     cin >> n;
 40     for (int i = 0; i < n; ++i) {
 41         scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
 42         edge[a1][b1] = edge[a2][b2] = true;    //边界上有线段顶点
 43         A[i].x = a1, A[i].y = b1, B[i].x = a2, B[i].y = b2;
 44     }
 45     scanf("%lf%lf", &T.x, &T.y);
 46     for (int i = 0; i < 100; ++i) {    //底边
 47         if (i && !edge[i][0]) {
 48             int cnt = 0;
 49             Point p(i, 0);
 50             for (int j = 0; j < n; ++j)
 51                 if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 52             ans = min(cnt, ans);
 53         }
 54         int cnt = 0;
 55         Point p(i + 0.5, 0);
 56         for (int j = 0; j < n; ++j)
 57             if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 58         ans = min(cnt, ans);
 59     }
 60     for (int i = 0; i < 100; ++i) {    //上边
 61         if (i && !edge[i][100]) {
 62             int cnt = 0;
 63             Point p(i, 100);
 64             for (int j = 0; j < n; ++j)
 65                 if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 66             ans = min(cnt, ans);
 67         }
 68         int cnt = 0;
 69         Point p(i + 0.5, 100);
 70         for (int j = 0; j < n; ++j)
 71             if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 72         ans = min(cnt, ans);
 73     }
 74     for (int i = 0; i < 100; ++i) {    //左侧边
 75         if (i && !edge[0][i]) {
 76             int cnt = 0;
 77             Point p(0, i);
 78             for (int j = 0; j < n; ++j)
 79                 if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 80             ans = min(cnt, ans);
 81         }
 82         int cnt = 0;
 83         Point p(0, i + 0.5);
 84         for (int j = 0; j < n; ++j)
 85             if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 86         ans = min(cnt, ans);
 87     }
 88     for (int i = 0; i < 100; ++i) {    //右侧边
 89         if (i && !edge[100][i]) {
 90             int cnt = 0;
 91             Point p(100, i);
 92             for (int j = 0; j < n; ++j)
 93                 if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
 94             ans = min(cnt, ans);
 95         }
 96         int cnt = 0;
 97         Point p(100, i + 0.5);
 98         for (int j = 0; j < n; ++j)
 99             if (SegmentProperIntersection(T, p, A[j], B[j]))++cnt;
100         ans = min(cnt, ans);
101     }
102     printf("Number of doors = %d\n", ans + 1);    //加上最外层边的那个 1
103 }

 

Treasure Hunt

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6995		Accepted: 2895

Description

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: 

Input

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).

Output

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

Sample Input

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 

Sample Output

Number of doors = 2