Submission #444810

#	Time	Username	Problem	Language	Result	Execution time	Memory
444810		zaneyu	Dango Maker (JOI18_dango_maker)	C++14	13 / 100	1 ms	392 KiB

dango_maker

/*input
5 5
RGRGW
GRRGW
WGGWR
RWRGW
RGWGW
*/
#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;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=double;
using pii=pair<int,int>;
#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(int 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 int 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)
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=3e3+5;
const int maxlg=__lg(maxn)+2;
string arr[maxn];
string ans="RGW";
bool hv[maxn][maxn];
int32_t main(){
    int n,m;
    cin>>n>>m;
    REP(i,n) cin>>arr[i];
    int cnt=0;
    REP(i,n){
        REP(j,m-2){
            if(hv[i][j]) continue;
            string a;
            REP(k,3){
                a.pb(arr[i][j+k]);
            }
            if(a==ans){
                REP(k,3){
                    hv[i][j+k]=1;
                }
                ++cnt;
            }
        }
    }
    REP(i,n-2){
        REP(j,m){
            bool die=0;
            string a;
            REP(k,3){
                if(hv[i+k][j]) die=1;
                a.pb(arr[i+k][j]);
            }
            if(die) continue;
            if(a==ans){
                REP(k,3) hv[i+k][j]=1;
                ++cnt;
            }
        }
    }
    cout<<cnt;
}  

• Subtask 1: 13 / 13
• Subtask 2: 0 / 20
• Subtask 3: 0 / 67