poj 3494 dp+单调栈

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 5088		Accepted: 1897
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 nnumbers. 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

求最大子矩阵（每一位都是1）

枚举每一个行，dp[n]前n行的最大子矩阵，如果最大子矩阵没用到最后一行则dp[n] = dp[n-1]。 如果用到第n行的话，因为矩阵必须是连续的，可以转换为poj 2559中n个宽都为相同的1，高度不同求最大面积问题，用单调栈的做法就可以仅用O(n)复杂度，总复杂度就是O(n^2)

#include 
#include 
#include 
#include 

using namespace std;

#define maxn 2001

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

int main()
{
    while(~scanf("%d %d", &m, &n)){
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
        for(int j = 1; j <= n; j++)
            for(int i = 2; i <= m; i++)
            if(a[i-1][j] && a[i][j])
                a[i][j] += a[i-1][j];   //求出每一列从第i行的位置上去共有几个连续的1


        stack s;
        int ans = 0;
        for(int i = 1; i <= m; i++){
            while(!s.empty()) s.pop();
            for(int j = 1; j <= n; j++){    //求第j个数的左边界
                while(!s.empty() && a[i][s.top()] >= a[i][j]) s.pop();
                l[j] = s.empty() ? 1 : s.top()+1;
                s.push(j);
            }
            while(!s.empty()) s.pop();
            for(int j = n; j >= 1; j--){    //求第j个数的右边界
                while(!s.empty() && a[i][s.top()] >= a[i][j]) s.pop();
                r[j] = s.empty() ? n : s.top()-1;
                s.push(j);
            }

            for(int j = 1; j <= n; j++)
                ans = max(ans, (r[j]-l[j]+1)*a[i][j]);
        }

        printf("%d\n", ans);


    }
    return 0;
}