spoj MINSUB 单调栈+二分

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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

题意：给你一个n*m的矩阵，求以一个最小值为m的最大矩阵面积s需要大于等于K

思路:二分答案，check怎么写呢。。

　　 类似bzoj 3039这题；

　　利用单调栈，求最大子矩阵面积；

　　 将check的x,大于等于x的值均改成1，求1的最大面积；

　　 枚举每个位置，以该位置能最大的上升的位置为权值；

　　 例如：

　　 1 0 1            1  0  1

　　 1 1 0    -->   2  1   0 

　　 1 0 1            3  0  1

　　 利用单调栈查找 以该权值为最大值最多可以往左和往右延伸最大长度；

　　 详见代码

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e3+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
int a[N][N],b[N][N];
int l[N],r[N],s[N];
int dp[N][N];
int n,m,k;
int check(int x)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        b[i][j]=(a[i][j]>=x);
    memset(dp,0,sizeof(dp));
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        if(b[i][j])dp[i][j]=dp[i-1][j]+1;
        else dp[i][j]=0;
        dp[i][0]=dp[i][m+1]=-1;
        int si=0;
        s[++si]=0;
        for(int j=1;j<=n;j++)
        {
            while(dp[i][s[si]]>=dp[i][j])si--;
            l[j]=s[si];
            s[++si]=j;
        }
        si=0;
        s[++si]=m+1;
        for(int j=m;j>=1;j--)
        {
            while(dp[i][s[si]]>=dp[i][j])si--;
            r[j]=s[si];
            s[++si]=j;
        }
        for(int j=1;j<=m;j++)
            ans=max(ans,(r[j]-l[j]-1)*dp[i][j]);
    }
    return ans;
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        int st=0;
        int en=1e9+10,ans=-1;
        while(st<=en)
        {
            int mid=(st+en)>>1;
            if(check(mid)>=k)
            {
                ans=mid;
                st=mid+1;
            }
            else
                en=mid-1;
        }
        printf("%d %d
",ans,check(ans));
    }
    return 0;
}