Submission #445118

#TimeUsernameProblemLanguageResultExecution timeMemory
445118zaneyuDango Maker (JOI18_dango_maker)C++14
100 / 100
992 ms100676 KiB
/*input
3 4
RGWR
GRGG
RGWW

*/
#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=4e3+7;
const int maxlg=__lg(maxn)+2;
string arr[maxn];
string ans="RGW";
int dp[maxn][maxn][2];
bool ok(int o,int i,int j){
    string a="";
    if(o) REP(k,3) a.pb(arr[i][j+k-1]);
    else REP(k,3) a.pb(arr[i+k-1][j]);
    return (a==ans);
}
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,m;
    cin>>n>>m;
    REP1(i,n){
        cin>>arr[i];
        arr[i]='0'+arr[i];
        arr[i].pb('0');
    }
    REP(i,n+1) REP(j,m+1) REP(a,2) dp[i][j][a]=-INF;
    int ans=0;
    for(int i=2;i<=n+m;i++){
        int mx=0;
        int X=-1,Y=-1;
        for(int x=max(1,i-m);x<=min(n,i-1);x++){
            int y=i-x;
            REP(a,2){
                if(ok(a,x,y)){
                    dp[x][y][a]=max(mx,dp[x-1][y+1][a])+1;
                }
            }
            REP(a,2) MXTO(mx,dp[x-1][y+1][a]);
            X=x,Y=y;
        }
        REP(a,2) MXTO(mx,dp[X][Y][a]);
        ans+=mx;
    }
    cout<<ans<<'\n';
}  
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