POJ1066Treasure Hunt【判断直线相交】

Language:Default Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5942   Accepted: 2466 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

题意：在金字塔内有一个宝藏p(x,y);现在要取出这个宝藏在金字塔内有许多墙为了进入宝藏所在的房间必须把墙炸开但是炸墙只能炸每个房间墙的中点求将宝藏运出城堡所需要的最小炸墙数；

解题思路：枚举每道墙的两个端点和p的连线这条线段和墙的交点的次数最小值即为需要炸墙的最小次数。

注意当墙数为零时输出1；

#include<cstdio>
#include<cstdlib>
#include<cstring>
#define eps 1e-6
#define inf 0x3f3f3f3f
using namespace std;
int MINI(int a,int b){
return a<b?a:b;
}
double MIN(double a,double b){
return a<b?a:b;
}
double MAX(double a,double b){
return a>b?a:b;
}
struct point{
double x,y;
};
struct line{
point a,b;
}A[35];
bool judge(int a,int b){
if(MIN(A[a].a.x,A[a].b.x)>MAX(A[b].a.x,A[b].b.x)||MIN(A[a].a.y,A[a].b.y)>MAX(A[b].a.y,A[b].b.y)||MIN(A[b].a.x,A[b].b.x)>MAX(A[a].a.x,A[a].b.x)||MIN(A[b].a.y,A[b].b.y)>MAX(A[a].a.y,A[a].b.y))
return false;
double h,i,j,k;
h=(A[a].b.x-A[a].a.x)*(A[b].a.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].a.x-A[a].a.x);
i=(A[a].b.x-A[a].a.x)*(A[b].b.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].b.x-A[a].a.x);
j=(A[b].b.x-A[b].a.x)*(A[a].a.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].a.x-A[b].a.x);
k=(A[b].b.x-A[b].a.x)*(A[a].b.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].b.x-A[b].a.x);
return h*i<=eps&&j*k<=eps;
}
int main()
{
int n,i,j,k,ans=inf;
double sx,sy;
scanf("%d",&n);
for(i=0;i<n;++i){
scanf("%lf%lf%lf%lf",&A[i].a.x,&A[i].a.y,&A[i].b.x,&A[i].b.y);
}
scanf("%lf%lf",&sx,&sy);
if(n==0){
ans=1;
printf("Number of doors = %d\n",ans);
return 0;
}
for(i=0;i<n;++i){
A
.a.x=sx;A
.a.y=sy;
A
.b.x=A[i].a.x;A
.b.y=A[i].a.y;
int cnt=0;
for(j=0;j<n;++j){
if(judge(j,n))cnt++;
}
ans=MINI(cnt,ans);
A
.b.x=A[i].b.x;A
.b.y=A[i].b.y;
cnt=0;
for(j=0;j<n;++j){
if(judge(j,n))cnt++;
}
ans=MINI(cnt,ans);
}
printf("Number of doors = %d\n",ans);
return 0;
}