제출 #1240890

#제출 시각아이디문제언어결과실행 시간메모리
1240890lioowOrchard (NOI14_orchard)C++20
25 / 25
83 ms23880 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #define pb push_back #define int long long #define repp(i,x,n) for(int i=x;i<=n;i++) #define rep(i,x,n) for(int i=x;i>=n;i--) #define cy cout<<"YES"<<endl #define cn cout<<"NO"<<endl #define ca cout<<"Alice"<<endl #define r0 return 0 #define fi first #define se second #define liow ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) #define jelek cout<<"jelek"<<endl #define pii pair<int,int> #define all(v) v.begin(),v.end() #define pi pair<pair<int,int>,pair<int,int>> #define p3 pair<int,pair<int,int>> #define p2 #define tp tuple<int,int,int> #define fl fflush(stdout) #define ld long double #define p5 pair<int,pair<pair<int,int>,pair<int,int>>> #pragma GCC optimize ("O2") #pragma GCC optimize ("unroll-loops") using namespace std; using namespace __gnu_pbds; typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set; typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset; typedef long long bryson; const int mod=1e9+7; const int SQMAX=635,INF=1e18; //const int mod=998244353; //const int MOD=1e6+3; mt19937_64 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count()); pii dr[8]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}}; int dx[4]={1,0,-1,0}; int dy[4]={0,1,0,-1}; int dxx[4]={1,1,-1,-1}; int dyy[4]={1,-1,-1,1}; int mul(int x,int y){return (x%mod*y%mod)%mod;} int expo(int b,int e){ if(e==0) return 1; int tmp=expo(b,e/2); if(e%2) return mul(tmp,mul(tmp,b)); else return mul(tmp,tmp); } signed main(){ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); int n,m;cin>>n>>m; int a[n+2][m+2]; memset(a,0,sizeof a); int satu=0; repp(i,1,n){ repp(j,1,m){ char x;cin>>x; if(x=='1') a[i][j]=1; else a[i][j]=-1; if(a[i][j]==1) satu++; if(i>1) a[i][j]+=a[i-1][j]; } } int maks=0,nw=0,banyak=0,brp=0; repp(l,1,n){ repp(i,l,n){ repp(j,1,m){ nw+=a[i][j]-a[i-l][j]; brp+=l; if(nw<0){ nw=0,brp=0; } if(maks<nw){ maks=nw; banyak=brp; } if(j==m) brp=nw=0; } } } int ans=satu-maks; // cout<<satu<<" "<<banyak<<" "<<maks<<endl; cout<<ans<<endl; }
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