POJ

题目链接：http://poj.org/problem?id=3494点击打开链接

Largest Submatrix of All 1’s

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 6674 Accepted: 2472Case Time Limit: 2000MS

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

Sample Output

04

单调栈的题 将矩阵里的每个1从上往下看做是一个长方形 则长度比他高的可以继续 比他小的停止

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;int mmap[2222][2222];int step[2222][2222];int l[2222][2222];int r[2222][2222];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        memset(mmap,0,sizeof(mmap));        memset(step,0,sizeof(mmap));        memset(l,0,sizeof(mmap));        memset(r,0,sizeof(mmap));        for(int i=1;i<=n;i++)            {                for(int j=1;j<=m;j++)                    scanf("%d",&mmap[i][j]);            }        int len=1;        for(int j=1;j<=m;j++)            {                for(int i=1;i<=n;i++)                {                    if(mmap[i][j])                    {                        step[i][j]=len++;                    }                    else                    {                        len=1;                    }                }                len=1;            }        stack<int > s;        for(int i=1;i<=n;i++)            {                for(int j=1;j<=m;j++)                   {                            while(1)                            {                                if(s.empty())                                    {                                        if(mmap[i][j]==0)                                            l[i][j]=0;                                        else                                            l[i][j]=1;                                        s.push(j);                                        break;                                    }                                    else                                    {                                        if(step[i][s.top()]>=step[i][j])                                        {                                            l[i][j]=s.top();                                            s.pop();                                        }                                        else                                        {                                            l[i][j]=s.top()+1;                                            s.push(j);                                            break;                                        }                                    }                            }                   }                   while(!s.empty())                            s.pop();            }            for(int i=1;i<=n;i++)            {                for(int j=m;j>=1;j--)                   {                            while(1)                            {                                if(s.empty())                                    {                                        if(mmap[i][j]==0)                                            r[i][j]=0;                                        else                                            r[i][j]=m;                                        s.push(j);                                        break;                                    }                                    else                                    {                                        if(step[i][s.top()]>=step[i][j])                                        {                                            r[i][j]=s.top();                                            s.pop();                                        }                                        else                                        {                                            r[i][j]=s.top()-1;                                            s.push(j);                                            break;                                        }                                    }                            }                   }                   while(!s.empty())                            s.pop();            }            int mmax=0;            for(int i=1;i<=n;i++)                for(int j=1;j<=m;j++)            {                mmax=max(mmax,(r[i][j]-l[i][j]+1)*step[i][j]);            }             printf("%d\n",mmax);            }}