奶酪
一个dfs就能完工......
从下往上,如果跑过一个洞就记录下来
如果这个洞跑不了,当我们返回的时候,再搜到这个洞,就跳过......
当我们找到出路后,可以安静等循环跑完或暴力跳出(怎么跳出我也不会)
然后得出答案......
(代码嫖的......之前思路错了,实在不想再写......见谅)
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int t,n,pd=0,vis[1005]; long long x[1005],y[1005],z[1005],h,r; double dis(int a,int b){ double x1=(double)x[a],y1=(double)y[a],z1=(double)z[a]; double x2=(double)x[b],y2=(double)y[b],z2=(double)z[b]; return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2)); } void search(int q){ if(z[q]+r>=h) { pd=1; return ;} vis[q]=1; for(int i=1;i<=n;i++){ if(!vis[i] and dis(q,i)<=2*r) search(i); } } int main(){ scanf("%d",&t); for(int i=1;i<=t;i++){ memset(x,-1,sizeof x); memset(y,-1,sizeof y); memset(z,-1,sizeof z); memset(vis,0,sizeof vis); scanf("%d%lld%lld",&n,&h,&r); for(int i=1;i<=n;i++){ scanf("%lld%lld%lld",&x[i],&y[i],&z[i]); } for(int i=1;i<=n;i++){ if(!vis[i] and z[i]-r<=0) search(i); if(pd==1){ cout<<"Yes"<<endl; break;} } if(pd==0) cout<<"No"<<endl; pd=0; } }