【BZOJ】【P3545】【ONTAK2010】【Peaks】【题解】【离线+并查集+平衡树启发式合并】

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3545

离线做法显然,并查集显然,Treap显然,启发式合并显然……加强版不会做唉……

Code:

#include<bits/stdc++.h>using namespace std;const int maxn=1e5+10;const int maxm=5e5+10;int n,m,q;int getint(){int res=0;char ch=getchar();while(!isdigit(ch))ch=getchar();while(isdigit(ch))res=(res<<3)+(res<<1)+ch-'0',ch=getchar();return res;}int rnd(){static int KEY=12345678;return KEY+=KEY<<2|1;}struct node{int val,key,size,s;node *c[2];    void rz(){          size=c[0]->size+s+c[1]->size;      }      void set(int _val=0,int _key=rnd(),int _size=1,int _s=1,node *C=NULL){          val=_val;key=_key;size=_size;s=_s;c[0]=c[1]=C;      }  };node pool[maxn];node* newnode(){static int tot=0;if(tot<maxn)return &pool[tot++];return new node();}struct Treap{node *root,*Null;Treap(){Null=newnode();Null->set(0,INT_MAX,0,0,Null);root=Null;}   void rot(node *&t,bool d){          node *p=t->c[d];t->c[d]=p->c[!d];          p->c[!d]=t;t->rz();p->rz();t=p;      }      void _insert(node *&t,int x){          if(t==Null){t=newnode();t->set(x,rnd(),1,1,Null);return;}          if(t->val==x){t->size+=1;t->s+=1;return;}          _insert(t->c[x>t->val],x);          if(t->c[x>t->val]->key<t->key)rot(t,x>t->val);          else t->rz();      }  int _kth(node *t,int x){        int r=t->c[0]->size;          if(t==Null)return 0;          else if(x<=r)return _kth(t->c[0],x);          else if(x>r+t->s)return _kth(t->c[1],x-r-t->s);          else return t->val;  }void insert(int x){_insert(root,x);}int kth(int x){return _kth(root,x);}int size(){return root->size;}}T[maxn];int fa[maxn];int find(int x){if(fa[x]!=x)return fa[x]=find(fa[x]);return x;}void Union(int x,int y){int fax=find(x),fay=find(y);if(fax==fay)return;if(T[fax].size()<T[fay].size())swap(x,y),swap(fax,fay);fa[fay]=fax;for(int i=0;i<T[fay].size();i++){int val=T[fay].kth(i+1);T[fax].insert(val);}}struct edge{int u,v,w;bool operator<(const edge &E)const{return w<E.w;}};edge edges[maxm];struct qes{int u,x,k,ind;bool operator<(const qes &q)const{return x<q.x;}};qes Q[maxm];int anss[maxm];int main(){n=getint();m=getint();q=getint();for(int i=1;i<=n;i++){int x=getint();T[i].insert(x);fa[i]=i;}for(int i=1;i<=m;i++){edges[i].u=getint();edges[i].v=getint();edges[i].w=getint();}sort(edges+1,edges+1+m);for(int i=1;i<=q;i++){Q[i].u=getint();Q[i].x=getint();Q[i].k=getint();Q[i].ind=i;}sort(Q+1,Q+1+q);int now=1;for(int i=1;i<=q;i++){for(;now<=m&&edges[now].w<=Q[i].x;now++)Union(edges[now].u,edges[now].v);int f=find(Q[i].u);Q[i].k=T[f].size()-Q[i].k;if(Q[i].k<0)anss[Q[i].ind]=-1;else anss[Q[i].ind]=T[f].kth(Q[i].k+1);}for(int i=1;i<=q;i++)printf("%d\n",anss[i]);return 0;}