Largest Submatrix of All 1’s--(单调队列)

Largest Submatrix of All 1’s

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 262144/131072K (Java/Other)
Total Submission(s) : 40   Accepted Submission(s) : 20
Problem Description

Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

 

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m lines each with n numbers. The input ends once EOF is met.

 

Output

For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.

 

Sample Input
 
   
2 2 0 0 0 0 4 4 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0
 

Sample Output
 
   
0 4

与之前题目:Largest Rectangle in a Histogram很像,题目是求最大矩形面积,与那题不同的的是,这题要以不同边为低分别求一次取最大值,题目先预处理使a[i][j]代表a[i][j]以及下面1的个数(相当于高度),然后对于a[i][j]再求出他左右比他连续高的列数(作为底)

代码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
int n,m,a[2005][2005],b[2005],q[2005],l[2005],r[2005];
ll work(int i)
{
    int j,tail=1; q[1]=0;
    for(j=1;j<=m;j++)
    {
        while(tail>=1&&a[i][q[tail]]>=a[i][j]) tail--;
        l[j]=j-q[tail]-1;
        q[++tail]=j;
    }
    tail=1; q[1]=n+1;
    for(j=m;j>=1;j--)
    {
        while(tail>=1&&a[i][q[tail]]>=a[i][j]) tail--;
        r[j]=q[tail]-1-j;
        q[++tail]=j;
    }
    ll maxn=0,temp=0;
    for(j=1;j<=m;j++)
    {
        temp=(l[j]+r[j]+1)*a[i][j];
        if(temp>maxn) maxn=temp;
    }
    return maxn;
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        ll ans=0,temp;
        for(i=n-1;i>=1;i--)
            for(j=1;j<=m;j++)
                if(a[i][j]!=0) a[i][j]=a[i+1][j]+1;
        for(i=1;i<=n;i++)
        {
            temp=work(i);
            if(temp>ans) ans=temp;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}


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