제출 #465887

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
465887		zaneyu	마상시합 토너먼트 (IOI12_tournament)	C++14	100 / 100	285 ms	34080 KiB

tournament

/*input
5 3 3
1 0 2 4
1 3
0 1
0 1
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
using pii=pair<int,int>;
using piii=pair<pii,int>;
typedef tree<piii,null_type,less<piii>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("O2","unroll-loops","no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;

#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=1e18+1;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-3;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
inline ll mult(ll a,ll b){
    if(a>=MOD) a%=MOD;
    if(b>=MOD) b%=MOD;
    return (a*b)%MOD;
}
inline ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=2e5+5;
const int maxlg=__lg(maxn)+2;
pii arr[maxn];
int par[maxn][20];
int table[maxn][20];
int query(int l,int r){
    int len=__lg(r-l+1);
    return max(table[l][len],table[r-(1<<len)+1][len]);
}
int GetBestPosition(int n, int c, int r, int *K, int *S, int *E) {
    indexed_set s;
    REP(i,n) s.insert({{i,i},i}),arr[i]={i,i},table[i][0]=K[i];
    int cur=n;
    REP(i,c){
        auto it=s.find_by_order(S[i]);
        int l=INF,r=0;
        REP(j,E[i]-S[i]+1){
            MNTO(l,(*it).f.f);
            MXTO(r,(*it).f.s);
            par[(*it).s][0]=cur;   
            it=next(it);
        }
        REP(j,E[i]-S[i]+1){
            s.erase(s.find_by_order(S[i]));
        }
        arr[cur]={l,r};
        s.insert({{l,r},cur++});
    }
    par[cur-1][0]=-1;
    REP1(j,19){
        REP(i,cur){
            if(par[i][j-1]!=-1){
                par[i][j]=par[par[i][j-1]][j-1];
            }
            else par[i][j]=-1;
        }
        REP(i,n){
            if(i+(1<<j)>n) break;
            table[i][j]=max(table[i][j-1],table[i+(1<<(j-1))][j-1]);
        }
    }
    int mx=-1,p=0;
    REP(i,n){
        int x=i;
        int ans=0;
        for(int j=19;j>=0;j--){
            int z=par[x][j];
            if(z!=-1 and query(arr[z].f,arr[z].s-1)<r){
                ans+=(1<<j);
                x=par[x][j];
            }
        }
        if(mx<ans) p=i,mx=ans;
    }
    return p;
}

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