H - Second Large Rectangle cattle off the second field training game

H - Second Large Rectangle

 

 

 

 

 

Topic analysis : to give you an N * M of 01 matrix allows you to find the second largest contains are all "1" size of the matrix. It requires O (n ^ 2) method to find the second largest matrix

The second large matrix there are two cases: 1. The second large full 1 matrix   2. The maximum of the largest sub-matrix of a matrix

Example: 1.

0111

0111

1111

1111

1111

1111     Its answer is 16   

2.

111111

111111 It answer is 2 * 5 rectangle 10

 

So we only need to build one answer sequence, the sequence where the size of the rectangle and the size of the largest of its largest sub-matrix, then the answer is sort sequence, the second largest is the answer

So how O (n ^ 2) to find the maximum matrix:

我们需要对遍历到的当前行的每个元素记录一个h[j],记录到这一行的这一列向上有多少个连续的1。然后记录一个l[j] , r[j]分别记录这个元素可以构成的矩形向左，向右最多可以到哪里。如图：高度为h[j]的矩形只能使用比它更高的。处理l[],r[]的复杂度为O(m)

 

 

 

代码:

#include <bits/stdc++.h>

using namespace std;
int c[1006][1006];
int l[1006],r[1006],h[1006];
int n,m;
int tot,p[4006000];
char str[1006];
set<pair<int,int> > s[1005];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%s",str+1);
        for(int j=1;j<=m;j++){
            c[i][j]=str[j]-'0';
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){  //h[]
            if(c[i][j])h[j]++;
            else h[j]=0;
        }
        for(int j=1;j<=m;j++){ //l[]
            l[j]=j;
            while(l[j]>1&&h[j]<=h[l[j]-1]){
                l[j]=l[l[j]-1];
            }
        }
        for(int j=m;j>=1;j--){ //r[]
            r[j]=j;
            while(r[j]<m&&h[j]<=h[r[j]+1]){
                r[j]=r[r[j]+1];
            }
        }
        for(int j=1;j<=m;j++){
            if(s[i].find(make_pair(l[j],r[j]))==s[i].end()){
                p[++tot] = h[j]*(r[j]-l[j]+1);
                s[i].insert(make_pair(l[j],r[j]));
                if(r[j]-l[j]+1 > h[j]){  //如果矩形的横边 大于竖边 那么它最大的子矩阵是（横边-1）*竖边
                    p[++tot] = h[j] * (r[j] - l[j]);
                }
            }
        }
    }
    sort(p+1,p+1+tot);
    if(tot<=1)printf("0\n");
    else printf("%d\n",p[tot-1]);
    return 0;
}