K: 奶酪

时间限制: 1 Sec  内存限制: 128 MB
提交: 115  解决: 30
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题目描述

输入

输出

样例输入

3

2 4 1

0 0 1

0 0 3

2 5 1

0 0 1

0 0 4

2 5 2

0 0 2

2 0 4

样例输出

Yes

No

Yes

提示

【题意】

题意很明确  能否从下面到上面

【思路】

转换方向， 就是一个图，   类似于 网络流  有一个 源点 和汇点，  唯一不同的就是  图里面 没有权值，   这样  其实 也可以用并查集， 源点 汇点    看看 最后是否在一个队里， 如果在  那么 就yes  否则 就 no   spfa  好理解，  从源点看看 汇点 可不可达

【代码实现】

SPFA   头文件  什么的 可以忽略~

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define  FI(n) IO::read(n)
#define  Be IO::begin()using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e5+5;
const int MAXN=50005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};namespace IO {const int MT = 5e7;char buf[MT]; int c,sz;void begin(){c = 0;sz = fread(buf, 1, MT, stdin);//一次性输入}template<class T>inline bool read(T &t){while( c < sz && buf[c] != '-' && ( buf[c]<'0' || buf[c] >'9')) c++;if( c>=sz) return false;bool flag = 0; if( buf[c]== '-') flag = 1,c++;for( t=0; c<=sz && '0' <=buf[c] && buf[c] <= '9'; c++ ) t= t*10 + buf[c]-'0';if(flag) t=-t;return true;}
}
ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}struct node{ll x,y,z;
}a[MAXN];
struct EDGE{ll v,next;ll w;
}edge[MAXN];
ll head[MAXN],cot;
int st=0,ed=MAXN-5;
int vis[MAXN];
void init()
{cot=0;mem(head,-1);
}
void add(ll u,ll v)
{edge[++cot].v=v;edge[cot].next=head[u];head[u]=cot;edge[++cot].v=u;edge[cot].next=head[v];head[v]=cot;
}
ll dis(ll x,ll y)
{return (pow(a[x].x-a[y].x,2)+pow(a[x].y-a[y].y,2)+pow(a[x].z-a[y].z,2));
}
bool SPFA()
{mem(vis,0);queue<ll>Q;Q.push(st);vis[st]=1;while(!Q.empty()){ll u=Q.front();Q.pop();// vis[u]=0;for(int i=head[u];i!=-1;i=edge[i].next){if(!vis[edge[i].v]){// printf("nest is %d\n",edge[i].v);vis[edge[i].v]=1;Q.push(edge[i].v);}}}
}
int main()
{int T;ll n,h,r;cin>>T;while(T--){init();memset(a,0,sizeof(a));mem(edge,0);scanf("%lld %lld %lld",&n,&h,&r);for(ll i=1;i<=n;i++){scanf("%lld %lld %lld",&a[i].x,&a[i].y,&a[i].z);}for(ll i=1;i<=n;i++){if(r>=a[i].z)add(st,i);if(a[i].z+r>=h)add(i,ed);for(ll j=i+1;j<=n;j++){if(dis(i,j)<= (4*r*r)){//printf("+++ %lld   %lld \n",dis(i,j),2*r);add(i,j);}}}SPFA();if(vis[ed])printf("Yes\n");elseprintf("No\n");}return 0;
}

并查集

//#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define  FI(n) IO::read(n)
#define  Be IO::begin()using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e5+5;
const int MAXN=50005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};namespace IO {const int MT = 5e7;char buf[MT]; int c,sz;void begin(){c = 0;sz = fread(buf, 1, MT, stdin);//一次性输入}template<class T>inline bool read(T &t){while( c < sz && buf[c] != '-' && ( buf[c]<'0' || buf[c] >'9')) c++;if( c>=sz) return false;bool flag = 0; if( buf[c]== '-') flag = 1,c++;for( t=0; c<=sz && '0' <=buf[c] && buf[c] <= '9'; c++ ) t= t*10 + buf[c]-'0';if(flag) t=-t;return true;}
}
ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}int st=0,ed=MAXN-5;
ll h,n,r;
struct node{ll x,y,z;
}a[MAXN];
ll pre[MAXN];
ll finds(ll x)
{return pre[x]==x? x:(pre[x]=finds(pre[x]));
}
void join(ll x,ll y)
{ll fx=finds(x);ll fy=finds(y);//printf("%d is %d | %d is %d\n",x,fx,y,fy);if(fx!=fy)pre[fx]=fy;
}
void init()
{pre[st]=st;pre[ed]=ed;for(ll i=1;i<=n;i++){pre[i]=i;}
}
ll dis(int x,int y)
{return  pow(a[x].x-a[y].x,2)+pow(a[x].y-a[y].y,2)+pow(a[x].z-a[y].z,2);
}
int main()
{int T;cin>>T;while(T--){scanf("%lld %lld %lld",&n,&h,&r);init();for(int i=1;i<=n;i++){scanf("%lld %lld %lld",&a[i].x,&a[i].y,&a[i].z);if(r>=a[i].z){join(i,st);}if(r+a[i].z>=h){join(i,ed);}}for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){if(dis(i,j)<=(4*r*r)){join(i,j);}}}int cot=0;//printf("pres is %d \n",finds(ed));if(finds(st)==finds(ed) )//|| finds(ed)==st)printf("Yes\n");elseprintf("No\n");}return 0;
}

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