P3958 奶酪

传送门

思路：

　　模拟题。用并查集求出所有 “连通块” ，判断是否有 “连通块” 的最顶上和最下方都不小于奶酪的范围。

Code:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<stack>
#include<vector>
#include<queue>
#include<deque>
#include<map>
#include<set>
using namespace std;
#define lck_max(a,b) ((a)>(b)?(a):(b))
#define lck_min(a,b) ((a)<(b)?(a):(b))
#define INF 1e9+7
typedef long long LL;
const int maxn=1e3+;
LL T,n,h,r,up_cnt,down_cnt,fa[maxn];
bool flag;
struct hh
{
    LL x,y,z,up,down;
}t[maxn*maxn];
inline LL read()
{
    LL kr=,xs=;
    char ls;
    ls=getchar();
    while(!isdigit(ls))
    {
        if(!(ls^))
            kr=-;
        ls=getchar();
    }
    while(isdigit(ls))
    {
        xs=(xs<<)+(xs<<)+(ls^);
        ls=getchar();
    }
    return xs*kr;
}
inline LL find(LL u)
{
    return u==fa[u]?u:fa[u]=find(fa[u]);
}
inline double dis(LL u,LL v)
{
    return sqrt((t[u].x-t[v].x)*(t[u].x-t[v].x)+(t[u].y-t[v].y)*(t[u].y-t[v].y)+(t[u].z-t[v].z)*(t[u].z-t[v].z));
}
int main()
{
    freopen("cheese.in","r",stdin);
    freopen("cheese.out","w",stdout);
    T=read();
    while(T--)
    {
        flag=false,up_cnt=,down_cnt=;
        n=read();h=read();r=read();
        for(LL i=;i<=n;i++) fa[i]=i;
        for(LL i=;i<=n;i++)
        {
            t[i].x=read();t[i].y=read();t[i].z=read();
            if(t[i].z+r>=h) t[++up_cnt].up=i;
            if(t[i].z-r<=) t[++down_cnt].down=i;
            for(LL j=;j<i;j++)
            {
                if(dis(i,j)>*r) continue;
                LL r1=find(fa[i]),r2=find(fa[j]);
                if(r1!=r2) fa[r1]=r2;
            }
        }
        for(LL i=;i<=up_cnt;i++)
            for(LL j=;j<=down_cnt;j++)
            {
                LL r1=find(t[i].up),r2=find(t[j].down);
                if(r1==r2) {flag=true;break;}
            }
        if(!flag) printf("No\n");
        else printf("Yes\n");
    }
    fclose(stdin);
    fclose(stdout);
return ;
}