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

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

 

Output

<p>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.</p>

 

Sample Input

2 20 00 04 40 0 0 00 1 1 00 1 1 00 0 0 0

 

Sample Output

04

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<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define inf 0x7fffffffint 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;}