[POJ 1066]Treasure Hunt

Posted 2020-09-30 kakakakakaka

Treasure Hunt

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7217		Accepted: 2965

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：

一个正方形的墓葬内有n堵墙，每堵墙的两个顶点都在正方形的边界上，现在这些墙将墓葬分割成了很多小空间，

已知正方形内的一个点上存在宝藏，现在我们要在正方形的外面去得到宝藏，对于每个小空间，我们可以炸开它的任意一条边的中点，

现在给出每堵墙的两个节点的坐标和宝藏的坐标，问如果要得到宝藏，需要炸的墙数最少是多少。

题解：

一道很有意思的计算几何，也比较巧妙。

我们可以发现，其实边界上的出发点，要到达终点，所经过的边数必定是这个点到终点连线与线段的焦点数。因为每一条边都是横穿整个正方形的，所以无论怎么走，都不可能绕过这一条边，毕竟不能走出正方形。

然后就是枚举出发点了，本人一开始将正方形四条边上的点分别排序，然后求中点。这样做很麻烦，仔细一想可以发现，无论是不是从两点的中点开始走，都会到达一个同样的方块。所以枚举中点是没有必要的，直接枚举正方形边上的所有点就可以了。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int n,ans=2000000000;
double xx,yy;
struct Vector
{
    double x,y;
    Vector(){}
    Vector(double xx,double yy){x=xx;y=yy;}
    Vector operator+(Vector b)
    {
        Vector ans;
        ans.x=x+b.x;ans.y=y+b.y;
        return ans;
    }
    Vector operator-(Vector b)
    {
        Vector ans;
        ans.x=x-b.x;ans.y=y-b.y;
        return ans;
    }
    double operator*(Vector b)
    {
        return x*b.y-b.x*y;
    }
    double operator^(Vector b)
    {
        return x*b.x+y*b.y;
    }
};
struct node
{
    Vector a,b;
}s[31],t;
bool check(double x,double y,int i)
{
    t.a=Vector(x,y);t.b=Vector(xx,yy);
    if(((t.a-t.b)*(s[i].a-t.b))*((t.a-t.b)*(s[i].b-t.b))<0&&((s[i].a-s[i].b)*(t.a-s[i].b))*((s[i].a-s[i].b)*(t.b-s[i].b))<0)return 1;
    else return 0;
}
void solve(double x,double y)
{
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(check(x,y,i))cnt++;
    }
    ans=min(ans,cnt);
}
int main()
{
    int i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%lf%lf%lf%lf",&s[i].a.x,&s[i].a.y,&s[i].b.x,&s[i].b.y);
    scanf("%lf%lf",&xx,&yy);
    for(i=0;i<=100;i++)
    {
        solve(i,100);
        solve(i,0);
        solve(0,i);
        solve(100,i);
    }
    printf("Number of doors = %d\n",++ans);
    return 0;
}