Submission #536076

#	Time	Username	Problem	Language	Result	Execution time	Memory
536076		zaneyu	Sandcastle 2 (JOI22_ho_t5)	C++14	10 / 100	5068 ms	596 KiB

Main

/*input
3 5
83 47 36 38 40
13 10 26 68 67
15 19 20 70 90
*/
#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("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;
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=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)
ll sub(ll a,ll b){
    ll x=a-b;
    while(x<0) x+=MOD;
    while(x>MOD) x-=MOD;
    return x;
}
ll mult(ll a,ll b){
    return (a*b)%MOD;
}
ll mypow(ll a,ll b){
    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;
}
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)?" ":"\n");
    return out;
}
const ll maxn=2e3+5;
const ll maxlg=__lg(maxn)+2; 
int arr[maxn];
bool vis[maxn];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,m;
    cin>>n>>m;
    REP(i,n) REP(j,m) cin>>arr[i*m+j];
    int ans=0;
    REP(i,n){
        REP(j,m){
            REP(a,i+1){
                REP(b,j+1){
                    pii p={a,b};
                    for(int x=a;x<=i;x++){
                        for(int y=b;y<=j;y++){
                            if(arr[p.f*m+p.s]<arr[x*m+y]) p={x,y};
                        }
                    }
                    bool wk=1;
                    REP(z,(i-a+1)*(j-b+1)-1){
                        int pp=-1;
                        REP(k,4){
                            int x=p.f+dx[k],y=p.s+dy[k];
                            if(x<a or x>i or y<b or y>j) continue;
                            if(arr[x*m+y]<arr[p.f*m+p.s] and (pp==-1 or arr[x*m+y]>arr[pp])) pp=x*m+y;
                        }
                        if(pp==-1){
                            wk=0;
                            break;
                        }
                        p={pp/m,pp%m};
                        //cout<<pp<<'\n';
                    }
                    //cout<<wk<<' '<<i<<' '<<a<<' '<<j<<' '<<b<<'\n';
                    ans+=wk;
                }
            }
        }
    }
    cout<<ans;
}

• Subtask 1: 0 / 9
• Subtask 2: 10 / 10
• Subtask 3: 0 / 5
• Subtask 4: 0 / 56
• Subtask 5: 0 / 20