[BZOJ3545][ONTAK2010][平衡树][STL][启发式合并]Peaks

STL tree练习题

两个坑点

• BZOJ上编译器版本较低，定义tree的时候null_type要改成null_mapped_type
• 讲道理splay_tree_tag要快一点，毕竟总复杂度是nlogn然而在BZOJ一直T，改成rb_tree_tag就过了……据说ov_tree_tag也过不了

其他的就跟普通的平衡树启发式合并一样做就好了

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <ctime>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define N 100010

using namespace std;
using namespace __gnu_pbds;

typedef pair<int,int> paris;

tree<paris,null_type,greater<paris>,rb_tree_tag,tree_order_statistics_node_update> T[N];
tree<paris,null_type,greater<paris>,rb_tree_tag,tree_order_statistics_node_update>::iterator itr;
//tree<paris,null_mapped_type,greater<paris>,rb_tree_tag,tree_order_statistics_node_update> T[N];
//tree<paris,null_mapped_type,greater<paris>,rb_tree_tag,tree_order_statistics_node_update>::iterator itr;  --BZOJ

int n,m,q;
int fa[N],Ans[N*6],Size[N];
paris tal[N];
struct stp{
  int x,y,w,g;
  friend bool operator <(stp a,stp b){
    return a.w<b.w;
  }
}E[N*6],Q[N*6];

inline char nc(){
  static char buf[100000],*p1=buf,*p2=buf;
  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}

inline void reaD(int &x){
  char c=nc(); x=0; int f=1;
  for(;c>57||c<48;c=nc())if(c=='-') f=-1;
  for(;c>=48&&c<=57;x=x*10+c-48,c=nc()); x*=f;
}

int find(int x){
  return x==fa[x]?x:fa[x]=find(fa[x]);
}

inline void Insert(int x,int y){
  x=find(x); y=find(y);
  if(x==y) return ;
  if(Size[x]>Size[y]) swap(x,y);
  for(itr=T[x].begin();itr!=T[x].end();itr++)
    T[y].insert(*itr);
  fa[x]=y;
  Size[y]+=Size[x];
  T[x].clear();
}

inline int query(int x,int y){
  x=find(x);
  if(Size[x]<y) return -1;
  return T[x].find_by_order(y-1)->first;
}

int w[30],wt;

inline void PutAns(int x){
  if(x==0){putchar(48);putchar('\n');return ;}
  if(x<0) putchar('-'),x=-x;
  while(x) w[++wt]=x%10,x/=10;
  for(;wt;wt--) putchar(w[wt]+48); putchar('\n');
}

int main(){
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  reaD(n); reaD(m); reaD(q);
  for(int i=1,x;i<=n;i++)
    reaD(x),tal[i]=paris(x,i),fa[i]=i,T[i].insert(tal[i]),Size[i]=1;
  for(int i=1;i<=m;i++)
    reaD(E[i].x),reaD(E[i].y),reaD(E[i].w);
  sort(E+1,E+1+m);
  for(int i=1;i<=q;i++)
    reaD(Q[i].x),reaD(Q[i].w),reaD(Q[i].y),Q[i].g=i;
  sort(Q+1,Q+1+q);
  for(int i=1,j=1;i<=q;i++){
    while(E[j].w<=Q[i].w&&j<=m)
      Insert(E[j].x,E[j].y),j++;
    Ans[Q[i].g]=query(Q[i].x,Q[i].y);
  }
  for(int i=1;i<=q;i++) PutAns(Ans[i]);
}