Submission #544968

#	Time	Username	Problem	Language	Result	Execution time	Memory
544968		zaneyu	Land of the Rainbow Gold (APIO17_rainbow)	C++14	11 / 100	3075 ms	14804 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=1e5+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;
}
vector<int> v[maxn],v2[maxn];
vector<int> bad[maxn];
int l=INF,r=0;
bool vis[2005][2005];
bool ok[2005][2005];
int n,m;
void init(int N, int C, int x, int y, int M, char *S) {
    //v[x].pb(y);
    n=N,m=C;
    MNTO(l,y),MXTO(r,y);
    ok[x][y]=1;
    REP(i,M){
        if(S[i]=='N') --x;
        else if(S[i]=='S') ++x;
        else if(S[i]=='E') ++y;
        else --y;
        //v[x].pb(y);
        MNTO(l,y),MXTO(r,y);
        ok[x][y]=1;
    }
    /*
    REP1(i,n){
        SORT_UNIQUE(v[i]);
        bad[i]=v[i];
        vector<int> vv;
        REP(j,sz(v[i])){
            int p=j;
            while(j<sz(v[i]) and v[i][j]-v[i][p]==j-p) ++j;
            --j;
            v2[i].pb(v[i][j]);
            vv.pb(v[i][p]);
        }
        //cout<<vv<<'\n';
        v[i]=vv;
    }*/
}
bool is(int x,int y){
    return ok[x][y];
}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int colour(int ar, int ac, int br, int bc) {
    int cnt=0;
    for(int i=ar;i<=br;i++){
        for(int j=ac;j<=bc;j++){
            if(!vis[i][j] and !is(i,j)){
                vis[i][j]=1;
                queue<pii> q;
                q.push({i,j});
                while(sz(q)){
                    pii z=q.front();
                    q.pop();
                    REP(k,4){
                        int x=z.f+dx[k],y=z.s+dy[k];
                        if(x<ar or x>br or y<ac or y>bc or is(x,y) or vis[x][y]) continue;
                        vis[x][y]=1;
                        q.push({x,y});
                    }
                }
                //cout<<i<<' '<<j<<'\n';
                ++cnt;
            }
        }
        //cout<<ans<<' ';
    }
for(int i=ar;i<=br;i++){
        for(int j=ac;j<=bc;j++){
            vis[i][j]=0;
        }
    }
    return cnt;
}

• Subtask 1: 11 / 11
• Subtask 2: 0 / 12
• Subtask 3: 0 / 24
• Subtask 4: 0 / 27
• Subtask 5: 0 / 26