Submission #531059

#TimeUsernameProblemLanguageResultExecution timeMemory
531059browntoadThe Kingdom of JOIOI (JOI17_joioi)C++14
100 / 100
2068 ms66100 KiB
#include <bits/stdc++.h> #pragma GCC optimize ("Ofast", "unroll-loops") using namespace std; #define ll long long #define int ll #define FOR(i,a,b) for (int i = (a); i<(b); i++) #define REP(i,n) FOR(i,0,n) #define REP1(i,n) FOR(i,1,n+1) #define RREP(i,n) for (int i=(n)-1; i>=0; i--) #define f first #define s second #define pb push_back #define ALL(x) x.begin(),x.end() #define SZ(x) (int)(x.size()) #define SQ(x) (x)*(x) #define pii pair<int, int> #define pdd pair<double ,double> #define pcc pair<char, char> #define endl '\n' //#define TOAD #ifdef TOAD #define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl #define IOS() #else #define bug(...) #define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0) #endif const ll inf = (1ll<<60); const int iinf=2147483647; const ll mod = 1e9+7; const ll maxn=2005; const double PI=acos(-1); ll pw(ll x, ll p, ll m=mod){ ll ret=1; while (p>0){ if (p&1){ ret*=x; ret%=m; } x*=x; x%=m; p>>=1; } return ret; } ll inv(ll a, ll m=mod){ return pw(a,m-2); } int a[2][maxn][maxn]; signed main(){ IOS(); int h,w; cin>>h>>w; int fmx=0, fmn=inf; REP(i,h){ REP(j,w){ cin>>a[0][i][j]; fmx=max(fmx, a[0][i][j]); fmn=min(fmn, a[0][i][j]); a[1][i][w-j-1]=a[0][i][j]; } } /*REP(id, 4){ if (id%2) swap(h, w); REP(i,h){ REP(j,w){ cout<<a[id][i][j]<<' '; } cout<<endl; } if (id%2) swap(h, w); cout<<endl; } */ int ans=inf; FOR(id, 0, 2){ int l=0, r=fmx-fmn; while(l<=r){ if (l==r){ ans=min(ans, l); break; } int mid=(l+r)/2; bool gg=false; int prevd=h; REP(i,w){ int k=-1; while(k+1<h&&a[id][k+1][i]<=fmn+mid){ k++; } k=min(k, prevd); prevd=k; //cout<<k<<endl; //cout<<acmx<<' '<<acmn<<endl; FOR(j, k+1, h){ if (fmx-mid>a[id][j][i]){ gg=1; break; } } if (gg) break; } if (gg){ l=mid+1; } else { r=mid; } } l=0; r=fmx-fmn; while(l<=r){ if (l==r){ ans=min(ans, l); break; } int mid=(l+r)/2; bool gg=false; int prevd=h; REP(i,w){ int k=-1; while(k+1<h&&a[id][k+1][i]>=fmx-mid){ k++; } k=min(k, prevd); prevd=k; //cout<<k<<endl; //cout<<acmx<<' '<<acmn<<endl; FOR(j, k+1, h){ if (mid+fmn<a[id][j][i]){ gg=1; break; } } if (gg) break; } if (gg){ l=mid+1; } else { r=mid; } } } cout<<ans<<endl; }
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