poj1066Treasure Hunt（线段相交）

链接

很纠结的找到了所有线段的中点，又很纠结的找到了哪些中点可以直接相连，最后bfs一下求出了最短路。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 2210
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double inf = ~0u>>;
 vector<int>ed[N];
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y(y) {} //构造函数 方便代码编写
 } p[],q[N];
 int g,dis[N];
 bool vis[N];
 vector<Point>pi[N];
 typedef Point pointt;
 pointt operator + (Point a,Point b)
 {
     return Point(a.x+b.x,a.y+b.y);
 }
 pointt operator - (Point a,Point b)
 {
     return Point(a.x-b.x,a.y-b.y);
 }
 pointt operator * (Point a,double b)
 {
     return Point(a.x*b,a.y*b);
 }
 pointt operator / (Point a,double b)
 {
     return Point(a.x/b,a.y/b);
 }
 bool operator < (const Point &a,const Point &b)
 {
     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 double cross(Point a,Point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 Point intersection1(Point a,Point b,Point c,Point d)//直线交点求解
 {
     Point pp = a;
     double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/
                ((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
     pp.x+=(b.x-a.x)*t;
     pp.y+=(b.y-a.y)*t;
     return pp;
 }
 bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
 {
     double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
                                         c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
     return dcmp(c1)*dcmp(c2)<&&dcmp(c3)*dcmp(c4)<;
 }
 void init(int i,int n)
 {
     p[i+].x = ;
     p[i+].y = ;
     p[i+].x = ;
     p[i+].y = ;
     p[i+].x = ;
     p[i+].y = ;
     p[i+].x = ;
     p[i+].y = ;
     p[i++n] = p[i+];
     p[i++n] = p[i+];
     p[i++n] = p[i+];
     p[i++n] = p[i+];

 }
 int judge(int x)
 {
     if(fabs(q[x].x)<eps||fabs(q[x].y)<eps||fabs(q[x].x-100.0)<eps||fabs(q[x].y-100.0)<eps)
         return ;
     return ;
 }
 void bfs(int st)
 {
     queue<int>Q;
     Q.push(st);
     memset(vis,,sizeof(vis));
    // cut<<
     for(int i =  ; i <= st ; i++)
         dis[i] = INF;
     dis[st] = ;
     while(!Q.empty())
     {
         int u = Q.front();
         Q.pop();
         //printf("%.2f %.2f %d\n",q[u].x,q[u].y,dis[u]);
         if(judge(u))
         {
             printf("Number of doors = %d\n",dis[u]);
             break;
         }
         for(int i = ; i < ed[u].size() ; i++)
         {
             int v = ed[u][i];
             //cout<<v<<" "<<dis[u]<<endl;
             dis[v] = min(dis[v],dis[u]+);
             if(!vis[v])
             {
                 vis[v] = ;
                 Q.push(v);
             }
         }
     }
 }
 bool cmp(Point a,Point b)
 {
     if(fabs(a.x-b.x)<eps)
         return a.y<b.y;
     return a.x<b.x;
 }
 int main()
 {
     int n,i,j,e;
     scanf("%d",&n);
 //    while(scanf("%d",&n)!=EOF)
 //    {
         n+=;
         g = ;
         for(i = ; i <= n- ; i++)
         {
             scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y);
         }
         init(n-,n);
         for(i = ; i <= n-; i++)
         {
             for(j = ; j <= n- ; j++)
             {
                 if(i==j) continue;
                 if(seginter(p[i],p[i+n],p[j],p[j+n]))
                 {
                     pi[i].push_back(intersection1(p[i],p[i+n],p[j],p[j+n]));
                 }
             }
         }
         for(i =  ; i <= n- ; i++)
         {
             if(dcmp(p[i].x)==)
                 pi[n-].push_back(p[i]);
             if(dcmp(p[i+n].x)==)
                 pi[n-].push_back(p[i+n]);
             if(dcmp(p[i].y-)==)
                 pi[n-].push_back(p[i]);
             if(dcmp(p[i+n].y-)==)
                 pi[n-].push_back(p[i+n]);
             if(dcmp(p[i].x-)==)
                 pi[n-].push_back(p[i]);
             if(dcmp(p[i+n].x-)==)
                 pi[n-].push_back(p[i+n]);
             if(dcmp(p[i].y)==)
                 pi[n].push_back(p[i]);
             if(dcmp(p[i+n].y)==)
                 pi[n].push_back(p[i+n]);
         }
         for(i =  ; i <= n; i++)
         {
             pi[i].push_back(p[i]);
             pi[i].push_back(p[i+n]);
             sort(pi[i].begin(),pi[i].end(),cmp);
             for(j =  ; j < pi[i].size()- ; j++)
             {
                 Point tp;
                 Point p1 = pi[i][j],p2= pi[i][j+];
                 if(dcmp(p1.x-p2.x)==&&dcmp(p1.y-p2.y)==) continue;
                 tp.x = (p1.x+p2.x)/;
                 tp.y = (p1.y+p2.y)/;
                 // printf("%d tpx = %.2f tpy = %.2f p1x = %.2f p1y = %.2f p2x = %.2f p2y = %.2f\n",i,tp.x,tp.y,p1.x,p1.y,p2.x,p2.y);
                 q[++g] = tp;
             }
             pi[i].clear();
         }
         g++;
         scanf("%lf%lf",&q[g].x,&q[g].y);
         for(i = ; i <= g ; i++)
             ed[i].clear();
         for(i = ; i <= g ; i++)
         {
             //  printf("%.2f %.2f\n",q[i].x,q[i].y);
             for(j = ; j <= g ; j++)
             {
                 if(i==j) continue;
                 for(e = ; e <= n; e++)
                 {
                     if(seginter(q[i],q[j],p[e],p[e+n]))
                     {
 //                        if(i==g)
 //                        printf("%.2f %.2f e = %d\n",q[j].x,q[j].y,e);
                         break;
                     }
                 }
                 if(e>n)
                 {
 //                    if(i==g)
 //                    printf("%.2f %.2f %.2f %.2f\n",q[i].x,q[i].y,q[j].x,q[j].y);
                     ed[i].push_back(j);
                 }
             }
         }
         bfs(g);
 //    }
     return ;
 }

感觉这题没那么复杂，交对后搜了下题解，，然后 这题是一个水题，确实水。。

因为这个点的最终目的是要走出去的，边界上墙的点相对于边界上其它的点应该是更优的，所以只需枚举这些点到达起点经过多少堵墙即可。

几何抓不住重点是件很纠结的事。

附优美的简短的跑得飞快的代码

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 2210
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double inf = ~0u>>;
 vector<int>ed[N];
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y(y) {} //构造函数 方便代码编写
 } p[],q[N];
 int g,dis[N];
 bool vis[N];
 vector<Point>pi[N];
 typedef Point pointt;
 pointt operator + (Point a,Point b)
 {
     return Point(a.x+b.x,a.y+b.y);
 }
 pointt operator - (Point a,Point b)
 {
     return Point(a.x-b.x,a.y-b.y);
 }
 pointt operator * (Point a,double b)
 {
     return Point(a.x*b,a.y*b);
 }
 pointt operator / (Point a,double b)
 {
     return Point(a.x/b,a.y/b);
 }
 bool operator < (const Point &a,const Point &b)
 {
     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 double cross(Point a,Point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
 {
     double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
                                         c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
     return dcmp(c1)*dcmp(c2)<&&dcmp(c3)*dcmp(c4)<;
 }
 int main()
 {
     int n,i,j,e;
     scanf("%d",&n);
     g = ;
     for(i = ; i <= n ; i++)
     {
         scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y);
     }
     Point tp;
     scanf("%lf%lf",&tp.x,&tp.y);
     if(n==)
     {
         puts("Number of doors = 1");
         return ;
     }
     int ans = INF;
     for(i =  ; i <= n*; i++)
     {
         int ts = ;
         for(j =  ; j <= n ; j++)
         {
             if(seginter(p[i],tp,p[j],p[j+n]))
                 ts++;
         }
         ans = min(ans,ts);
     }
     printf("Number of doors = %d\n",ans);
     return ;
 }