标签：就是 最大 space 最小 通过 影响 直接 一条直线 strong
已知平面内 N 个点的坐标，求欧氏距离下的第 K 远点对。
输出文件第一行为一个整数，表示第 K 远点对的距离的平方（一定是个整数）。
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N=100010; 5 int n,m,WD,mn[N][2],mx[N][2],ch[N][2],rt; 6 struct P{ 7 int x,y; 8 P(int _x=0,int _y=0):x(_x),y(_y){}; 9 P operator -(const P&a)const{return P(x-a.x,y-a.y);} 10 bool operator <(const P&a)const{return WD?y<a.y:x<a.x;} 11 }p[N],q; 12 priority_queue<ll,vector<ll>,greater<ll> >ans; 13 ll len(P a){return (ll)a.x*a.x+(ll)a.y*a.y;} 14 void build(int&k,int l,int r,int d){ 15 k=(l+r)>>1; 16 WD=d;nth_element(p+l,p+k,p+r+1); 17 mn[k][0]=mx[k][0]=p[k].x; 18 mn[k][1]=mx[k][1]=p[k].y; 19 if(l<k){build(ch[k][0],l,k-1,d^1);} 20 if(k<r){build(ch[k][1],k+1,r,d^1);} 21 for(int i=0;i<2;++i)if(ch[k][i]){ 22 int t = ch[k][i]; 23 mn[k][0]=min(mn[t][0],mn[k][0]); 24 mn[k][1]=min(mn[t][1],mn[k][1]); 25 mx[k][0]=max(mx[t][0],mx[k][0]); 26 mx[k][1]=max(mx[t][1],mx[k][1]); 27 } 28 } 29 inline ll sqr(int x){return (ll)x*x;} 30 inline ll cal(int k){ 31 if(!k)return 0; 32 return max(sqr(q.x-mn[k][0]),sqr(q.x-mx[k][0])) + max(sqr(q.y-mn[k][1]),sqr(q.y-mx[k][1])) ; 33 } 34 void query(int k){ 35 ll tmp = len(p[k]-q); 36 // printf("%lld\n",tmp); 37 if(tmp>ans.top())ans.pop(),ans.push(tmp); 38 ll tl = cal(ch[k][0]), tr = cal(ch[k][1]); 39 if(tl>tr){ 40 if(tl>ans.top())query(ch[k][0]); 41 if(tr>ans.top())query(ch[k][1]); 42 }else{ 43 if(tr>ans.top())query(ch[k][1]); 44 if(tl>ans.top())query(ch[k][0]); 45 } 46 } 47 int main(){ 48 #ifndef ONLINE_JUDGE 49 freopen("T2.in","r",stdin); 50 freopen("T2.out","w",stdout); 51 #endif 52 scanf("%d%d",&n,&m); 53 for(int i=1;i<=n;++i)scanf("%d%d",&p[i].x,&p[i].y); 54 for(int i=1;i<=m*2;++i)ans.push(0); 55 build(rt,1,n,0); 56 for(int i=1;i<=n;++i){q=p[i];query(rt);} 57 cout<<ans.top()<<endl; 58 return 0; 59 }
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N=100010; 5 int n,k,vis[N],id[N]; 6 priority_queue<ll,vector<ll>,greater<ll> >ans; 7 char gc(){ 8 static char*p1,*p2,s[1000000]; 9 if(p1==p2)p2=(p1=s)+fread(s,1,1000000,stdin); 10 return(p1==p2)?EOF:*p1++; 11 } 12 int rd(){ 13 int x=0; char c=gc(); 14 while(c<‘0‘||c>‘9‘)c=gc(); 15 while(c>=‘0‘&&c<=‘9‘)x=(x<<1)+(x<<3)+c-‘0‘,c=gc(); 16 return x; 17 } 18 struct P{ 19 int x,y; 20 P(int _x=0,int _y=0):x(_x),y(_y){}; 21 P operator -(const P&a)const{return P(x-a.x,y-a.y);} 22 bool operator <(const P&a)const{return x==a.x?y<a.y:x<a.x;} 23 }p[N],q[N]; 24 inline ll len(P a){return (ll)a.x*a.x+(ll)a.y*a.y;} 25 inline ll crs(P a,P b){return (ll)a.x*b.y-(ll)a.y*b.x;} 26 void ins(ll x){ 27 if((int)ans.size()<k)ans.push(x); 28 else { 29 if(x<=ans.top())return; 30 ans.pop();ans.push(x); 31 } 32 } 33 bool solve(){ 34 int top=1,i; 35 i=1;while(vis[i])++i; 36 q[top=1]=p[i];id[top]=i; 37 for(++i;i<=n;++i)if(!vis[i]){ 38 while(top>1&&crs(q[top]-q[top-1],p[i]-q[top])<=0)top--; 39 q[++top]=p[i],id[top]=i; 40 } 41 int now=top; 42 i=n;while(vis[i])--i; 43 for(--i;i;--i)if(!vis[i]){ 44 while(top>now&&crs(q[top]-q[top-1],p[i]-q[top])<=0)top--; 45 q[++top]=p[i],id[top]=i; 46 } 47 top--; 48 if(top==2){ 49 vis[id[1]]=vis[id[2]]=1; 50 ins(len(q[1]-q[2])); 51 } 52 int x = 2;ll mx=-1,pos1,pos2; 53 for(i=1;i<=top;++i){ 54 P A = q[i%top+1]-q[i]; 55 while(x!=i&&crs(A,q[x%top+1]-q[x])>0)x=x%top+1; 56 ll t = len(q[x]-q[i]); 57 if(t>mx)mx=t,pos1=id[i],pos2=id[x]; 58 } 59 if(len(p[pos1]-p[pos2])<=ans.top())return false; 60 for(i=1;i<=n;++i)if(!vis[i]&&i!=pos1){ 61 ins(len(p[i]-p[pos1])); 62 } 63 for(i=1;i<=n;++i)if(!vis[i]&&i!=pos1&&i!=pos2){ 64 ins(len(p[i]-p[pos2])); 65 } 66 vis[pos1]=vis[pos2]=1; 67 return true; 68 } 69 int main(){ 70 freopen("T2.in","r",stdin); 71 freopen("T2.out","w",stdout); 72 n=rd(),k=rd(); 73 for(int i=1;i<=n;++i)p[i].x=rd(),p[i].y=rd(); 74 sort(p+1,p+n+1); 75 for(int i=1;i<=k;++i)ans.push(0); 76 for(int i=1;i<=min(k,n-1);++i)if(!solve())break; 77 cout<<ans.top()<<endl; 78 return 0; 79 }
标签：就是 最大 space 最小 通过 影响 直接 一条直线 strong
原文地址：https://www.cnblogs.com/Paul-Guderian/p/10306228.html
