제출 #261031

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
261031		uacoder123	마상시합 토너먼트 (IOI12_tournament)	C++14	100 / 100	278 ms	12400 KiB

tournament

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define F first
#define S second
#define FOR(i,a,b) for (auto i = (a); i <= (b); ++i)
#define NFOR(i,a,b) for(auto i = (a); i >= (b); --i)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
#define mp(i,a) make_pair(i,a)
#define pb(a) push_back(a)
#define bit(x,b) (x&(1LL<<b))

typedef int lli;
typedef pair <lli,lli> ii;
typedef pair <lli,ii> iii;
typedef vector <lli> vi;
typedef tree<ii,null_type,less<ii>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

int n,c,r;
int segt[400001],segt1[400001],lazy[400001];
void up(int node,int l,int r,int in,int v)
{
  if(l==r)
  {
    segt[node]=v;
    return;
  }
  int m=(l+r)/2;
  if(in<=m)
    up(2*node+1,l,m,in,v);
  else
    up(2*node+2,m+1,r,in,v);
  segt[node]=max(segt[2*node+1],segt[2*node+2]);
}
int qu(int node,int l,int r,int s,int e)
{
  if(l>e||r<s)
    return(0);
  if(l>=s&&r<=e)
    return segt[node];
  int m=(l+r)/2;
  int q1=qu(2*node+1,l,m,s,e),q2=qu(2*node+2,m+1,r,s,e);
  return max(q1,q2);
}
void up1(int node,int l,int r,int s,int e)
{
  if(l>e||r<s)
    return;
  if(lazy[node]!=0)
  {
    segt1[node]+=lazy[node];
    if(l!=r)
    {
      lazy[2*node+1]+=lazy[node];
      lazy[2*node+2]+=lazy[node];
    }
    lazy[node]=0;
  }
  if(l>=s&&r<=e)
  {
    segt1[node]+=1;
    if(l!=r)
    {
      lazy[2*node+1]+=1;
      lazy[2*node+2]+=1;  
    }
    return;
  }
  int m=(l+r)/2;
  up1(2*node+1,l,m,s,e);
  up1(2*node+2,m+1,r,s,e);
  segt1[node]=max(segt1[2*node+1],segt1[2*node+2]);
}
int qu1(int node,int l,int r,int s,int e)
{
  if(l>e||r<s)
    return(0);
  if(lazy[node]!=0)
  {
    segt1[node]+=lazy[node];
    if(l!=r)
    {
      lazy[2*node+1]+=lazy[node];
      lazy[2*node+2]+=lazy[node];
    }
    lazy[node]=0;
  }
  if(l>=s&&r<=e)
    return segt1[node];
  int m=(l+r)/2;
  int q1=qu1(2*node+1,l,m,s,e),q2=qu1(2*node+2,m+1,r,s,e);
  return max(q1,q2);
}
ordered_set s;
vector<ii> v;
int GetBestPosition(int N, int C, int R, int *K, int *S, int *E) 
{
  n=N;
  c=C;
  r=R;
  for(int i=0;i<n-1;++i)
    up(0,0,n-2,i,K[i]);
  for(int i=0;i<n;++i)
    s.insert(mp(i,i));
  for(int i=0;i<c;++i)
  {
    int f=S[i],sec=E[i],co=0,fi,se;
    while(co<(sec-f))
    {
      if(co==0)
        fi=(*s.find_by_order(f)).S;
      s.erase((*s.find_by_order(f)));
      co++;
    }
    se=(*s.find_by_order(f)).F;
    s.erase((*s.find_by_order(f)));
    v.pb(mp(fi,se));
    s.insert(mp(se,fi));
  }
  for(int i=0;i<c;++i)
  {
    if(qu(0,0,n-2,v[i].F,v[i].S-1)<r)
      up1(0,0,n-1,v[i].F,v[i].S);
  }
  int m=0,in=0,cur;
  for(int i=0;i<n;++i)
  {
    cur=qu1(0,0,n-1,i,i);
    if(cur>m)
    {
      m=cur;
      in=i;
    }
  }
  return in;
}

• Subtask 1: 17 / 17
• Subtask 2: 32 / 32
• Subtask 3: 51 / 51