题目给一个01矩阵,求最大的1子矩阵。

先用dp预处理出每一行的每一列的1能向上按连续的1延伸多少,然后枚举每一行作为子矩阵的底,那样对于每一行的答案就是POJ2559这个经典问题了。

 #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 2222
int stack[MAXN],top,l[MAXN],r[MAXN];
int calc(int *a,int n){
a[++n]=-; top=;
for(int i=; i<=n; ++i){
l[i]=r[i]=i;
while(top && a[stack[top]]>a[i]){
r[stack[top]]=i-;
l[i]=l[stack[top]];
--top;
}
if(top && a[stack[top]]==a[i]) l[i]=l[stack[top]];
stack[++top]=i;
}
int res=;
for(int i=; i<n; ++i){
res=max(res,a[i]*(r[i]-l[i]+));
}
return res;
}
int d[MAXN][MAXN];
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
for(int i=; i<=n; ++i){
for(int j=; j<=m; ++j) scanf("%d",&d[i][j]);
}
for(int i=; i<=n; ++i){
for(int j=; j<=m; ++j){
if(d[i][j]) d[i][j]=d[i-][j]+;
}
}
int res=;
for(int i=; i<=n; ++i){
res=max(res,calc(&d[i][],m));
}
printf("%d\n",res);
}
return ;
}

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