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[1005];
bool v[1005];
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[1005],cnt;
struct node
{
    int to,nex;
}e[2005005];
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,0,sizeof(v));
    while(!q.empty())q.pop();
    q.push(0);v[0]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();if(x==n+1)return 1;
        for(int i=head[x];i;i=e[i].nex)
        {
            int y=e[i].to;
            if(v[y])continue;
            q.push(y);v[y]=1;
        }
    }
    return 0;
}
int main()
{
    //freopen("cheese.in","r",stdin);
    //freopen("cheese.out","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cnt=0;memset(head,0,sizeof(head));
        scanf("%d%d%d",&n,&h,&r);
        for(int i=1;i<=n;++i)
        {
            scanf("%lld%lld%lld",&p[i].x,&p[i].y,&p[i].z);
        }
        p[0].x=p[0].y=p[0].z=0;p[n+1].x=p[n+1].y=0;p[n+1].z=h;
        for(int i=1;i<=n;++i)
        {
            for(int j=i+1;j<=n;++j)
            {
                double dis=c(p[i],p[j]);
                if(dis>r+r)continue;add(i,j),add(j,i);
            }
        }
        for(int i=1;i<=n;++i)
        {
            if(p[i].z<=r&&p[i].z>=-r)add(0,i);
            if(p[i].z>=h-r&&p[i].z<=h+r)add(i,n+1);
        }
        if(bfs())puts("Yes");
        else puts("No");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}