POJ-1066-Treasure Hunt（线段相交）

题目链接：http://poj.org/problem?id=1066

题目大意：一个100*100的房间，里面有n个墙，题目保证不会有三个墙交于一个点。问离开房间的最少穿墙次数（包括房间的墙壁）

思路：n的范围很小，枚举所有顶点，然后找到最小的即可。

ACCode：

//#pragma comment(linker, "/STACK:1024000000,1024000000")
  
#include<stdio.h>
#include<string.h> 
#include<math.h> 
   
#include<map>  
#include<set>
#include<deque> 
#include<queue> 
#include<stack> 
#include<bitset>
#include<string> 
#include<fstream>
#include<iostream> 
#include<algorithm> 
using namespace std; 
  
#define ll long long 
#define Pair pair<int,int>
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// ??
//std::ios::sync_with_stdio(false);
//  register
const int MAXN=1e2+10;
const int INF32=0x3f3f3f3f;
const ll INF64=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
const double EPS=1.0e-8;

struct Point{
	double x,y,t,d;
	Point(double _x=0,double _y=0,double _t=0,double _d=0){
		x=_x;y=_y;t=_t;d=_d;
	}
	friend Point operator + (const Point &a,const Point &b){
		return Point(a.x+b.x,a.y+b.y);
	}
	friend Point operator - (const Point &a,const Point &b){
		return Point(a.x-b.x,a.y-b.y);
	}
	friend double operator ^ (Point a,Point b){//??????? 
		return a.x*b.y-a.y*b.x;
	}
	friend int operator == (const Point &a,const Point &b){
		if(fabs(a.x-b.x)<EPS&&fabs(a.y-b.y)<EPS) return 1;
		return 0;
	}
};
struct V{
	Point start,end;double ang;
	V(Point _start=Point(0,0),Point _end=Point(0,0),double _ang=0.0){
		start=_start;end=_end;ang=_ang;
	}
	friend V operator + (const V &a,const V &b){
		return V(a.start+b.start,a.end+b.end);
	}
	friend V operator - (const V &a,const V &b){
		return V(a.start-b.start,a.end-b.end);
	}
};
V Line[MAXN];
Point S;
int n;

double Distance(Point a,Point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int LineInter(V l1,V l2){
	if(max(l1.start.x,l1.end.x)>=min(l2.start.x,l2.end.x)&&
	max(l2.start.x,l2.end.x)>=min(l1.start.x,l1.end.x)&&
	max(l1.start.y,l1.end.y)>=min(l2.start.y,l2.end.y)&&
	max(l2.start.y,l2.end.y)>=min(l1.start.y,l1.end.y)){
		if(((l2.end-l2.start)^(l1.start-l2.start))*((l2.end-l2.start)^(l1.end-l2.start))<=0&&
		((l1.end-l1.start)^(l2.start-l1.start))*((l1.end-l1.start)^(l2.end-l1.start))<=0)
			//printf(" 相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
			return 1;
	}//printf("不相交： l1:(%lf,%lf),(%lf,%lf) l2:(%lf,%lf),(%lf,%lf)\n",l1.start.x,l1.start.y,l1.end.x,l1.end.y,l2.start.x,l2.start.y,l2.end.x,l2.end.y);
	return 0;
}
int main(){
	scanf("%d",&n);
	double x1,x2,y1,y2;
	for(int i=1;i<=n;++i){
		scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
		Line[i]=V(Point(x1,y1),Point(x2,y2));
	}scanf("%lf%lf",&S.x,&S.y);
	int ans=INF32;
	for(int i=1;i<=n;++i){
		int cnt=0;
		for(int j=1;j<=n;++j){
			if(LineInter(Line[j],V(S,Line[i].start))) cnt++;
		}ans=min(ans,cnt);
		cnt=0;
		for(int j=1;j<=n;++j){
			if(LineInter(Line[j],V(S,Line[i].end))) cnt++;
		}ans=min(ans,cnt);
	}printf("Number of doors = %d \n",ans==INF32?1:ans);
}