D. Maximum Submatrix 2
time limit per test
memory limit per test
input
output
You are given a matrix consisting of digits zero and one, its size is n × m. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?
Let's assume that the rows of matrix a are numbered from 1 to n from top to bottom and the columns are numbered from 1 to m from left to right. A matrix cell on the intersection of the i-th row and the j-th column can be represented as (i, j). Formally, a submatrix of matrix a is a group of four integers d, u, l, r (1 ≤ d ≤ u ≤ n; 1 ≤ l ≤ r ≤ m). We will assume that the submatrix contains cells (i, j)(d ≤ i ≤ u; l ≤ j ≤ r). The area of the submatrix is the number of cells it contains.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 5000). Next n lines contain m characters each — matrix a. Matrix a only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.
Output
Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.
Examples
input
1 1
1
output
1
input
2 2
10
11
output
2
input
4 3
100
011
000
101
output
2
用str[][]存储整个矩阵
d[i][j]表示str[i][j]为'1'的情况下，第i行以str[i][j]为结尾的连续的最长的1的长度(若str[i][j] == '1')d[i][j] = d[i][j-1] + 1
遍历d[][]的每一列执行操作vis[d[i][j]]++, 求每个长度的个数，遍历完一列后，从后到前执行vis[i-1] += vis[i]
更新ans = max(ans, vis[i]*i)
#include <bits/stdc++.h>
#define maxn 5005
#define MOD 1000000007
typedef long long ll;
using namespace std;
char str[maxn][maxn];
int d[maxn][maxn];
int vis[maxn];
int main() {
// freopen("in.txt", "r", stdin);
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
scanf("%s", str[i]+1);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++) {
if(str[i][j] == '1') {
d[i][j] = d[i][j-1] + 1;
}
}
int ans = 0;
for(int i = 1; i <= m; i++){
memset(vis, 0, sizeof(vis));
for(int j = 1; j <= n; j++) {
vis[d[j][i]]++;
}
for(int j = 5000; j >= 1; j--) {
vis[j-1] += vis[j];
ans = max(ans, vis[j] * j);
}
}
printf("%d\n", ans);
return 0;
}