NOIP2017 D2T1奶酪

这题终于是正经第一题感觉了。

只需要对相交或相切的球建一条边，然后对所有与底面有交点的球连边，再对所有与顶面有交点的球连边，bfs判断上下连通性即可。

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 using namespace std;
 typedef long long ll;
 struct point
 {
     long long x,y,z;
 }p[];
 bool v[];
 double c(point a,point b)
 {
     return sqrt(1ll*(a.x-b.x)*(a.x-b.x)+1ll*(a.y-b.y)*(a.y-b.y)+1ll*(a.z-b.z)*(a.z-b.z));
 }
 int head[],cnt;
 struct node
 {
     int to,nex;
 }e[];
 void add(int x,int y)
 {
     e[++cnt].to=y;e[cnt].nex=head[x];head[x]=cnt;
 }
 queue<int>q;int n,h,r;
 bool bfs()
 {
     memset(v,,sizeof(v));
     while(!q.empty())q.pop();
     q.push();v[]=;
     while(!q.empty())
     {
         int x=q.front();q.pop();if(x==n+)return ;
         for(int i=head[x];i;i=e[i].nex)
         {
             int y=e[i].to;
             if(v[y])continue;
             q.push(y);v[y]=;
         }
     }
     return ;
 }
 int main()
 {
     //freopen("cheese.in","r",stdin);
     //freopen("cheese.out","w",stdout);
     int t;
     scanf("%d",&t);
     while(t--)
     {
         cnt=;memset(head,,sizeof(head));
         scanf("%d%d%d",&n,&h,&r);
         for(int i=;i<=n;++i)
         {
             scanf("%lld%lld%lld",&p[i].x,&p[i].y,&p[i].z);
         }
         p[].x=p[].y=p[].z=;p[n+].x=p[n+].y=;p[n+].z=h;
         for(int i=;i<=n;++i)
         {
             for(int j=i+;j<=n;++j)
             {
                 double dis=c(p[i],p[j]);
                 if(dis>r+r)continue;add(i,j),add(j,i);
             }
         }
         for(int i=;i<=n;++i)
         {
             if(p[i].z<=r&&p[i].z>=-r)add(,i);
             if(p[i].z>=h-r&&p[i].z<=h+r)add(i,n+);
         }
         if(bfs())puts("Yes");
         else puts("No");
     }
     //fclose(stdin);
     //fclose(stdout);
     return ;
 }