牛客网（暑假集训#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矩阵中求第二大矩形。

Analysis

首先考虑求最大矩形：

不妨考虑以第i行为底边的矩形，那么就转化成直方图求最大矩形面积。

如图

当我们遇到第三列时，第二列的高度已经不能拓展，因此将其清算掉，同时将第三列高度加入栈中。

即使用单调栈维护所有高度中最靠右且小于当前高度的位置（注意单调）。

进而考虑求第二大，情况有二，所有完整的尽量大的矩形中第二大，或者尽量大的矩形去掉一行或者一列。由于我们按行以此求解，只需要额外考虑去掉一列的情况就OK。

另：可以使用滚动数组边读入边求解。

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

补题网址