Submission #545769

#	Time	Username	Problem	Language	Result	Execution time	Memory
545769		zaneyu	Land of the Rainbow Gold (APIO17_rainbow)	C++14	100 / 100	1135 ms	205644 KiB

rainbow

#include "rainbow.h"
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("O2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,ll>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int 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 maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#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)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
struct seg{
    set<int> s[maxn];
    int rt[maxn];
    int seg[20*maxn],lc[20*maxn],rc[20*maxn];
    int cur=2;
    void add(int x,int y){
        s[x].insert(y);
    }
    void build(){
        rt[0]=1;
        REP(i,maxn-1){
            for(auto x:s[i]) upd(rt[i],1,maxn,x);
            rt[i+1]=rt[i];
        }
    }
    void upd(int &idx,int l,int r,int p){
        seg[cur]=seg[idx]+1,lc[cur]=lc[idx],rc[cur]=rc[idx];
        idx=cur++;
        if(l==r){
            return;
        }
        int mid=(l+r)/2;
        if(p<=mid) upd(lc[idx],l,mid,p);
        else upd(rc[idx],mid+1,r,p);
    }
    int query(int l1,int r1,int l2,int r2){
        if(l2>r2) return 0;
        return query(rt[r1],1,maxn,l2,r2)-query(rt[l1-1],1,maxn,l2,r2);
    }
    int query(int idx,int l,int r,int ql,int qr){
        if(r<ql or l>qr) return 0;
        if(ql<=l and r<=qr) return seg[idx];
        int mid=(l+r)/2;
        return query(lc[idx],l,mid,ql,qr)+query(rc[idx],mid+1,r,ql,qr);
    }
}ver,edh,edv,rvrs;
int xmx=0,xmn=INF,ymx=0,ymn=INF;
void add(int x,int y){
    ver.add(x,y),ver.add(x+1,y+1),ver.add(x+1,y),ver.add(x,y+1);
    edh.add(x,y),edh.add(x+1,y);
    edv.add(x,y),edv.add(x,y+1);
    rvrs.add(x,y);
}
void init(int N, int C, int x, int y, int M, char *S) {
    xmx=xmn=x;
    ymx=ymn=y;
    add(x,y);
    REP(i,M){
        if(S[i]=='N') --x;
        else if(S[i]=='S') ++x;
        else if(S[i]=='E') ++y;
        else --y;
        add(x,y);
        MNTO(ymn,y),MXTO(ymx,y),MXTO(xmx,x),MNTO(xmn,x);
    }
    ver.build(),edh.build(),edv.build(),rvrs.build();
}
int colour(int ar, int ac, int br, int bc) {
    int e=edh.query(ar+1,br,ac,bc)+edv.query(ar,br,ac+1,bc);
    int v=ver.query(ar+1,br,ac+1,bc);
    int r=rvrs.query(ar,br,ac,bc);
    int c=2;
    if(ar>=xmn or br<=xmx or ac>=ymn or bc<=ymx) c=1;
    //cout<<e<<' '<<v<<' '<<r<<' '<<c<<'\n';
    return e-v+c-r;
}

• Subtask 1: 11 / 11
• Subtask 2: 12 / 12
• Subtask 3: 24 / 24
• Subtask 4: 27 / 27
• Subtask 5: 26 / 26