Largest Submatrix of All 1’s(POJ 3494) 单调栈

来自《挑战程序设计竞赛》

1.题目原文

http://poj.org/problem?id=3494

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 5885		Accepted: 2219
Case 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 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

Source

POJ Founder Monthly Contest – 2008.01.31, xfxyjwf

2.解题思路

求一个0-1矩阵中全为1的最大子矩阵。

关键是看出最大子矩阵，一定是从某个位置开始，向上连续1到达最高点，然后分别左右延伸得到最大面积，即为最大子矩阵种元素的个数。

因此，先预处理出所有点向上连续1的个数。然后利用单调栈，以这个数为高，每行都是O(n)的左右分别延伸求面积的最大值。这一步就转化为了挑战上的例题:求柱形图中矩形的最大面积。这题难点在于，是二维矩阵，也就是对每一行都用挑战例题中的方法求解。时间复杂度为O(n^2)。

挑战上的例题，见 http://blog.youkuaiyun.com/qq_33929112/article/details/52744490#t1

3.AC代码

#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<set>
#include<vector>
#include<cmath>
#include<bitset>
#include<stack>
#include<sstream>
#include<deque>
#include<utility>
using namespace std;
#define maxn 2005
typedef pair<int,int> P;

int m,n;
int a[maxn][maxn];
int u[maxn][maxn];
int l[maxn][maxn],r[maxn][maxn];

void pre()
{
    memset(u,0,sizeof(u));
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            if(a[i][j]==0) u[i][j]=0;
            else u[i][j]=u[i-1][j]+1;
        }
    }
}

void solve()
{
    stack<P> s;
    pre();
    for(int i=1;i<=m;i++){
        while(!s.empty()) s.pop();
        for(int j=1;j<=n;j++){
            while(!s.empty()&&s.top().first>=u[i][j]){
                s.pop();
            }
            if(s.empty()) l[i][j]=1;
            else l[i][j]=s.top().second+1;
            s.push(P(u[i][j],j));
        }
    }

    for(int i=1;i<=m;i++){
        while(!s.empty()) s.pop();
        for(int j=n;j>=1;j--){
            while(!s.empty()&&s.top().first>=u[i][j]){
                s.pop();
            }
            if(s.empty()) r[i][j]=m;
            else r[i][j]=s.top().second-1;
            s.push(P(u[i][j],j));
        }
    }

    int res=0;
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            res=max(res,u[i][j]*(r[i][j]-l[i][j]+1));
        }
    }
    printf("%d\n",res);
}

int main()
{
    while(scanf("%d%d",&m,&n)!=EOF){
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&a[i][j]);
            }
        }
        solve();
    }
    return 0;
}