CF376 D. Maximum Submatrix 2
阿新 • • 發佈:2021-07-01
題目傳送門:https://codeforces.com/problemset/problem/376/D
題目大意:
給你 \(n\times m\) 的01矩陣,問你在對行(Row)任意排序後,最大的全1子矩陣大小
因為每一行裡面的相對位置不會發生改變,故我們預處理一下
記\(R[i][j]\)表示位置\((i,j)\)能向右能延伸多遠
然後對於每個\(j\),我們按\(R[i][j]\)的值對\(1\sim n\)行排序
類似單調棧的思想往下找即可
/*program from Wolfycz*/ #include<map> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair<int,int> #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } template<typename T>inline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } template<typename T>inline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=5e3; int F[N+10][N+10]; char Map[N+10][N+10]; int main(){ // freopen(".in","r",stdin); // freopen(".out","w",stdout); int n=read(0),m=read(0),Ans=0; for (int i=1;i<=n;i++) scanf("%s",Map[i]+1); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) F[j][i]=!(Map[i][j]-'0')?0:F[j-1][i]+1; for (int j=1;j<=m;j++){ sort(F[j]+1,F[j]+1+n); reverse(F[j]+1,F[j]+1+n); for (int i=1;i<=n;i++){ if (!F[j][i]) break; Ans=max(Ans,F[j][i]*i); } } printf("%d\n",Ans); return 0; }