MINSUB - Largest Submatrix

时间：2021-03-26 17:05:49

MINSUB - Largest Submatrix

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You are given an matrix M (consisting of nonnegative integers) and an integer K.  For any submatrix of M' of M define min(M') to be the minimum value of all the entries of M'.  Now your task is simple:  find the maximum value of min(M') where M' is a submatrix of M of area at least K (where the area of a submatrix is equal to the number of rows times the number of columns it has).

Input

The first line contains a single integer T (T ≤ 10) denoting the number of test cases, T test cases follow.  Each test case starts with a line containing three integers, R (R ≤ 1000), C (C ≤ 1000) and K (K ≤ R * C) which represent the number of rows, columns of the matrix and the parameter K.  Then follow R lines each containing C nonnegative integers, representing the elements of the matrix M.  Each element of M is ≤ 10^9

Output

For each test case output two integers:  the maximum value of min(M'), where M' is a submatrix of M of area at least K, and the maximum area of a submatrix which attains the maximum value of min(M').  Output a single space between the two integers.

Example

Input:

2

2 2 2

1 1

1 1

3 3 2

1 2 3

4 5 6

7 8 9

Output:

1 4

8 2
分析：首先二分答案M，其次改写成01矩阵，这样变成求最大的全1子矩阵；
　　　最大全1子矩阵那么可以单调栈解决；
代码：
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <bitset>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define sys system("pause")

const int maxn=1e3+;

const int N=1e3+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t,qu[maxn],a[maxn][maxn],v[maxn][maxn],le[maxn],ret,ma,now;

bool ok(int x)

{

    now=;

    int i,j;

    rep(i,,n)rep(j,,m)v[i][j]=(a[i][j]>=x?v[i-][j]+:);

    rep(i,,n)

    {

        rep(j,,m+)

        {

            if(!qu[])qu[++qu[]]=v[i][j],le[qu[]]=j;

            else if(v[i][j]>qu[qu[]])qu[++qu[]]=v[i][j],le[qu[]]=j;

            else if(v[i][j]<qu[qu[]])

            {

                int tmp;

                while(qu[]>=&&v[i][j]<=qu[qu[]])

                {

                    now=max(now,(j-le[qu[]])*qu[qu[]]);

                    tmp=le[qu[]];

                    qu[]--;

                }

                qu[++qu[]]=v[i][j];

                le[qu[]]=tmp;

            }

        }

        qu[]--;

    }

    return now>=k;

}

int main()

{

    int i,j;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d%d",&n,&m,&k);

        rep(i,,n)rep(j,,m)scanf("%d",&a[i][j]);

        int l=,r=1e9;

        while(l<=r)

        {

            int mid=l+r>>;

            if(ok(mid))ret=mid,ma=now,l=mid+;

            else r=mid-;

        }

        printf("%d %d\n",ret,ma);

    }

    return ;

}

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