poj 1066 Treasure Hunt

http://poj.org/problem?id=1066

大意; 问在房间中有一份宝藏，但是房间中有一些隔板，问最少需要通过多少隔板
思路： 链接宝藏与爆破地点，枚举每一条直线寻找最少破坏的隔板。。 

 1 #include <iostream>
  2 #include <algorithm>
  3 #include <cmath>
  4 using namespace std;
  5 const int maxn = 110;
  6 const double eps = 1e-8;
  7 
  8 double fix_double(double x){
  9     return fabs(x)<eps?0.0:x;
 10 }
 11 
 12 int sign(double x){
 13     return x<-eps?-1:x>eps;
 14 }
 15 
 16 struct point {
 17     double x,y;
 18     //point (){}
 19     point(double x=0,double y=0):x(x),y(y){}
 20     point operator - (const point b) const {
 21         return point(x-b.x,y-b.y);
 22     }
 23 };
 24 
 25 struct line {
 26     point a,b;
 27     line(){}
 28     line(point _a,point _b){
 29         a = _a;
 30         b = _b;
 31     }
 32 };
 33 double dot(point o,point a,point b){
 34     point oa = a-o;
 35     point ob = b-o;
 36     return oa.x*ob.x+oa.y*ob.y;
 37 }
 38 
 39 double cross(point o,point a,point b){
 40     point oa = a-o;
 41     point ob = b-o;
 42     return oa.x*ob.y-ob.x*oa.y;
 43 }
 44 
 45 int n;
 46 line ln[maxn];
 47 point tr,pt[maxn*10];//tr为宝藏，pt为边上的点
 48 
 49 double dis(point a,point b){
 50     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 51 }
 52 
 53 bool cmp(point a,point b){  //数据太弱这样也过了。。。
 54     point zero = point(0,0);
 55     int ans = sign(cross(zero,a,b));
 56     if(ans==0) return dis(zero,a)+eps<dis(zero,b);
 57     return ans>0;
 58 }
 59 //0不在，1，在线段内，2在断点处
 60 int in_line(point pt, line li){
 61     if(sign(cross(pt,li.a,li.b))) return 0;
 62     int ans = sign(dot(pt,li.a,li.b));
 63     return ans==0?2:ans<0;//cos钝角小于0.。
 64 }
 65 
 66 //0不相交，1规范相交，2交与短点，3，有重合部分
 67 int across(line ab,line cd){
 68     if(max(ab.a.x,ab.b.x)<min(cd.a.x,cd.b.x)||max(ab.a.y,ab.b.y)<min(cd.a.y,cd.b.y)||max(cd.a.x,cd.b.x)<min(ab.a.x,ab.b.x)||max(cd.a.y,cd.b.y)<min(ab.a.y,ab.b.y))
 69         return 0;
 70     int ab_cd = sign(cross(ab.a,ab.b,cd.a))*sign(cross(ab.a,ab.b,cd.b));
 71     int cd_ab = sign(cross(cd.a,cd.b,ab.a))*sign(cross(cd.a,cd.b,ab.b));
 72     if(ab_cd==0&&cd_ab==0&&(in_line(ab.a,cd)==1||in_line(ab.b,cd)==1))
 73         return 3;
 74     if(ab_cd<0)
 75         return cd_ab==0?2:cd_ab<0;
 76     else if(ab_cd==0)
 77         return cd_ab<=0?2:0;
 78     return 0;
 79 }
 80 
 81 int solve(){
 82     int ans = 0x3f3f3f3f,cnt=0;
 83     for(int i=0;i<n;i++){
 84         pt[cnt++] = ln[i].a;
 85         pt[cnt++] = ln[i].b;
 86     }
 87     pt[cnt++] = point(0,0);
 88     pt[cnt++] = point(0,100);
 89     pt[cnt++] = point(100,0);
 90     pt[cnt++] = point(100,100);
 91     sort(pt,pt+cnt,cmp);
 92     for(int i=1;i<cnt;i++){
 93         point cn = point((pt[i].x+pt[i-1].x)/2,(pt[i].y+pt[i-1].y)/2);
 94         line lt = line(tr,cn);
 95         int tot=0;
 96         for(int j=0;j<n;j++){
 97             if(across(lt,ln[j])) tot++;
 98         }
 99         if(tot<ans) ans = tot;
100     }
101     return ans+1;
102 }
103 int main()
104 {
105     while(cin>>n){
106         for(int i=0;i<n;i++){
107             cin>>ln[i].a.x>>ln[i].a.y>>ln[i].b.x>>ln[i].b.y;
108         }
109         cin>>tr.x>>tr.y;
110         int res = solve();
111         cout<<"Number of doors = "<<res<<endl;
112     }
113     return 0;
114 }