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) : 41   Accepted Submission(s) : 21

Problem Description

Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? Bylargest 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

 

Source

PKU

 

题意：

给你一个矩阵，由0和1组成，让你求最大的一块矩阵。

思路：

一开始毫无头绪，但是后来仔细一想，只需要将矩阵进行压缩即可，注意：这里的压缩不是整列压缩，而是一行一行压缩，这就转化成了前面所做过的
最大面积的了。

代码：

#include<iostream>#include<cstring>#include<stdio.h>using namespace std;int main(){int t[2010];int h[2010];int l[2010];int r[2010];int s[2010];int n,m;int maxx;while(~scanf("%d%d",&n,&m) && n && m){    maxx=0;    memset(t,0,sizeof(t));    memset(h,0,sizeof(h));    memset(s,0,sizeof(s));    for(int k=1;k<=m;k++){    for(int j=1;j<=n;j++)    {        scanf("%d",&t[j]);    if(t[j]==0)h[j]=0;    else h[j]+=t[j];    l[j]=j;r[j]=j;    }int top=0;for(int i=1;i<=n;i++){    while(top && h[i]<=h[s[top]])top--;if(top==0)l[i]=1;else l[i]=s[top]+1;s[++top]=i;}top=0;for(int i=n;i>=1;i--){    while(top && h[i]<=h[s[top]])top--;            if(top==0)                r[i]=n;            else                r[i]=s[top]-1;            s[++top]=i;            }int maxn=0;int summ;for(int i=1;i<=n;i++){summ=h[i]*(r[i]-l[i]+1);if(maxn<summ)maxn=summ;}if(maxx<maxn)maxx=maxn;}printf("%d\n",maxx);}return 0;}

那个dp求最大矩阵和压缩方法为：

for(i=1;i<=n;i++){   for(j=1;j<=n;j++)b[j]=0;//初始化   for(j=i;j<=n;j++){       for(k=1;k<=n;k++)       b[k]+=a[j][k];        }.....}