Cattle passenger network (summer camp # 2) H - Second Large Rectangle

Description

Given a $\mathrm{N\times M binary\; matrix.\; Please\; output\; the\; size\; of\; second\; large\; rectangle\; containing\; all\; "1".\; Containing\; all "1" means\; that\; the\; entries\; of\; the\; rectangle\; are\; all "1".}$

A rectangle can be defined as four integers ${\mathrm{x1,y1,x2,y2 where\; 1\le x1\le x2\le N and\; 1\le y1\le y2\le M.\; Then,\; the\; rectangle\; is\; composed\; of\; all\; the\; cell\; (x,\; y)\; where\; x1\le x\le x2 and\; y1\le y\le y2.\; If\; all\; of\; the\; cell\; in\; the\; rectangle\; is\; "1",\; this\; is\; a\; valid\; rectangle.}}^{}$

Please find out the size of the second largest rectangle, two rectangles are different if exists a cell belonged to one of them but not belonged to the other.

Input

The first line of input contains two space-separated integers N and M.
Following N lines each contains M characters c_ij${\mathrm{.1\le N,M\le 1000}}^{}$

${\mathrm{N\times M\ge 2}}^{}$

${\mathrm{cij\in "01"}}^{}$

Output

Output one line containing an integer representing the answer. If there are less than 2 rectangles containning all $\text{"1", output "0".}$

Sample Input 1

1 2

01

 

Sample Output 1

0

 

Sample Input 2

1 3

101

 

Sample Output 2

1

 

Resume

01 seek the second largest rectangular matrix.

 

Analysis

First consider seeking the maximum rectangle:

Consider the rectangle to the bottom edge of the i-th behavior, then transformed into the largest rectangle area histogram requirements.

Figure

When we encounter the third column, the height of the second row can not expand, and therefore it is liquidating, while the third column at the height of the stack.

I.e. monotonically maintain all the stack height and the rightmost position of a height smaller than the current (note monotone).

Further consider second largest demand, there are two cases, all as large as possible a complete second largest rectangle, or as large rectangular or remove a row. Since we solved this row, you only need to remove the additional consideration of a case on OK.

Another: You can use the scroll reads the array side edge solving.

Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N = 1010;
 4 int n,m,m1,m2,h[N],s[N],top;
 5 bool gra[N][N];
 6 
 7 inline int Max(int a,int b){return a>b?a:b;}
 8 
 9 int main(){
10     scanf("%d %d",&n,&m);
11     for(int i=1;i<=n;i++){
12         getchar();
13         for(int j=1;j<=m;j++){
14             gra[i][j] = getchar()-'0';
15         }
16     }
17 
18     for(int i=1;i<=n;i++){
19         top = 0;
20         int ans;
21         for(int j=1;j<=m+1;j++){
22             h[j] = gra[i][j]?h[j]+1:0;
23             while(top && h[j] <= h[s[top]]){
24                 //printf("!%d %d\n",i,s[top-1]);
25                 ans = h[s[top]]*(j-1-s[top-1]);
26                 if(ans > m1){
27                     m2 = m1;
28                     m1 = ans;
29                     ans -= h[s[top]];
30                     if(ans > m2){
31                         m2 = ans;
32                     }
33                 }
34                 else if(ans > m2){
35                     m2 = ans;
36                 }
37                 
38                 top--;
39             }
40             s[++top] = j;
41         }
42     }
43 
44     printf("%d\n",m2);
45     return 0;
46 }

View Code

Appendix

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