Submission #638437

#	Time	Username	Problem	Language	Result	Execution time	Memory
638437		zaneyu	Maxcomp (info1cup18_maxcomp)	C++14	100 / 100	129 ms	16964 KiB

maxcomp

/*input
2 3
2 4 3
5 7 5
*/
#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;
//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()))))
#define GET(v,x) lower_bound(ALL(v),x)-v.begin()
const int maxn=1e3+5;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
const ld PI=acos(-1.0l);
const ld eps=1e-6;
const ll INF64=9e18+1;
#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)?"\n":"");
    return out;
}
int arr[maxn][maxn],pf[maxn][maxn];
int n,m;
bool isin(int x,int y){
    if(x<0 or x>=n or y<0 or y>=m) return false;
    return true;
}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int get(){
    int ans=-1;
    REP(i,n+1) REP(j,m+1) pf[i][j]=INF;
    REP1(i,n){
        REP1(j,m){
            int x=arr[i][j]-i-j;
            pf[i][j]=min({pf[i][j-1],pf[i-1][j],x});
            MXTO(ans,x-pf[i][j]-1);
        }
    }
    return ans;
}
int main(){
    ios::sync_with_stdio(false),cin.tie(0); 
    cin>>n>>m;
    vector<pair<int,pii>> v;
    REP1(i,n) REP1(j,m){
        cin>>arr[i][j];
    }
    int ans=get();
    REP1(i,n) REP1(j,m/2) swap(arr[i][j],arr[i][m-j+1]);
    MXTO(ans,get());
    REP1(i,n/2) REP1(j,m) swap(arr[i][j],arr[n-i+1][j]);
    MXTO(ans,get());
    REP1(i,n) REP1(j,m/2) swap(arr[i][j],arr[i][m-j+1]);
    MXTO(ans,get());
    cout<<ans;
}

• Subtask 1: 15 / 15
• Subtask 2: 15 / 15
• Subtask 3: 30 / 30
• Subtask 4: 40 / 40