uva 754 Treasure Hunt?! （线段位置判断并且求交点）

只需要枚举每个起点与终点之间有条线段与它相交。

代码：

    #include<cstdio>      #include<cstring>      #include<cstdlib>      #include<algorithm>      #define MAXN 1000      #define ESP 1e-8      using namespace std;      struct point       {          double x,y;          point(): x(.0), y(.0) {};          point(double _x,double _y) : x(_x),y(_y) {};          void input() { scanf("%lf %lf",&x,&y); }        };        struct line      {          point a,b;      };      double sgn(double d)      {   return d > ESP ? 1 : (d < -ESP ? -1 : 0);      }      point operator-(const point a,const point b)      {   return point(b.x-a.x , b.y-a.y);      }      double operator*(const point a,const point b)      {   return a.x*b.y - a.y*b.x;      }      bool operator==(const point a,const point b)      {   return sgn(a.x-b.x) == 0 && sgn(a.y-b.y) == 0;      }      bool jiaodian(point a,point b,point c,point d,point &p)      {          double s1 = (b-a) * (c-a) , s2 = (b-a) * (d-a),                 s3 = (d-c) * (a-c) , s4 = (d-c) * (b-c);          if( sgn(s1) * sgn(s2) > 0 || sgn(s3) * sgn(s4) > 0)             return false;          p = point( (s2*c.x - s1*d.x) / (s2-s1),                      (s2*c.y - s1*d.y) / (s2-s1) );          return true;      }      bool pan(point a,point b,point c,point d)      {          double s1 = (b-a) * (c-a) , s2 = (b-a) * (d-a),                 s3 = (d-c) * (a-c) , s4 = (d-c) * (b-c);          if( sgn(s1) * sgn(s2) >= 0 || sgn(s3) * sgn(s4) >= 0)              return false;          return true;      }      line ab[MAXN];      point d[MAXN],s;      int dn,n;      int max(int a, int b) { return a>b?a:b; }      int min(int a, int b) { return a<b?a:b; }      bool cmp(const point &a, const point &b)      {          if( sgn(a.x-b.x) != 0 )           {  if( sgn(a.x-b.x) < 0 ) return true;             else return false;          }          else           {  if( sgn(a.y-b.y) < 0)  return true;             else  return false;          }      }      int get(point a,point b)      {          int i,j,k = 0;          point c;          for(i=1;i<=n;i++)            if(pan(a, b, ab[i].a, ab[i].b))               k++;             return k;      }      int add(point a,point b)      {          int i,j,k,r,w,ans;          point c;          dn = 0;          d[++dn] = a;           d[++dn] = b;          for(i=1;i<=n;i++)          {  if( jiaodian(a, b, ab[i].a, ab[i].b, c) )                d[++dn] = c;          }          sort(d+1,d+dn+1,cmp);         // for(i=1;i<=dn;i++)           //   printf("%f %f \n",d[i].x,d[i].y);                              ans = 1e8;          for(i=2;i<=dn;i++)          {  k = get(s, point( (d[i-1].x+d[i].x)/2, (d[i-1].y+d[i].y)/2 ) ) ;             //printf("%f %f %d\n",(d[i-1].x+d[i].x)/2, (d[i-1].y+d[i].y)/2,k);              ans = min(ans, k);          }          return ans;      }      void init()      {          int i,j,k,r,w;                    //scanf("%d",&n);          for(i=1;i<=n;i++)          {  ab[i].a.input();             ab[i].b.input();          }           s.input();      }       void solve()      {          int ans = 1e8;          ans = min(ans, add(point(0,0), point(0,100)));          ans = min(ans, add(point(0,100), point(100,100)));          ans = min(ans, add(point(100,100), point(100,0)));          ans = min(ans, add(point(0,0), point(100,0)));          printf("Number of doors = %d\n",ans+1);      }      int main()      {          int t,i;          //scanf("%d",&t);          //for(i=1;i<=t;i++)          while(scanf("%d",&n) != EOF)          {  init();             solve();          }          return  0;      }