SPOJ-MINSUB

题目链接

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:22 2 21 11 13 3 21 2 34 5 67 8 9Output:1 48 2

 Submit solution!

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题意：

给定一个由非负数组成的矩阵M，和一个整数K，对于矩阵M的子矩阵M’，定义min(M’)为M'矩阵中元素的最小值。

我们需要找出这样一个子矩阵，该矩阵的面积至少为K，且min(M’)最大化。面积的定义为该矩阵的行数*列数。求出min(M'),并给出使得min(M')为该值时面积的最大值。

题解：

这类问题都是可以二分答案的。把小于二分值的位置设为0，其他设为1，那么问题就变成了求全为1的子矩阵的最大面积，这件事情可以用单调栈搞，可以预处理出每个点(i,j)的左边有多少个连续的1(记为f[i][j])，然后枚举每一列，那么这一列就可以类似与单调栈一样处理算出能上下延伸多少，做法和一位数组的单调栈一样。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <stack>using namespace std;const int maxn = 1e3 + 5;int maze[maxn][maxn], val[maxn][maxn], l[maxn];int n, m, k;int check(int x){    memset(val, 0, sizeof(val));    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)        {            if(maze[i][j] >= x) val[i][j] = 1;            if(val[i][j]) val[i][j] = val[i][j-1] + 1;        }    int ans = -1;    for(int j = 1; j <= m; j++)    {        stack<int> s;        val[n+1][j] = -1;        for(int i = 1; i <= n+1; i++)        {            while(s.size() && val[s.top()][j] >= val[i][j])            {                ans = max(ans, (i-l[s.top()])*val[s.top()][j]);                s.pop();            }            l[i] = s.size() == 0 ? 1 : s.top()+1;            s.push(i);        }    }    return ans;}int main(){    int t;    cin >> 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", &maze[i][j]);        int l = 0, r = 1e9, mid, ans;        while(l <= r)        {            mid = (l+r)/2;            if(check(mid) >= k)                ans = mid, l = mid + 1;            else                r = mid - 1;        }        printf("%d %d\n", ans, check(ans));    }    return 0;}