Second Large Rectangle

题目链接（点击）

题目描述

Given a N×MN \times MN×M binary matrix. Please output the size of second large rectangle containing all "1"\texttt{"1"}"1".

Containing all "1"\texttt{"1"}"1" means that the entries of the rectangle are all "1"\texttt{"1"}"1".

A rectangle can be defined as four integers x1,y1,x2,y2x_1, y_1, x_2, y_2x1​,y1​,x2​,y2​ where 1≤x1≤x2≤N1 \leq x_1 \leq x_2 \leq N1≤x1​≤x2​≤N and 1≤y1≤y2≤M1 \leq y_1 \leq y_2 \leq M1≤y1​≤y2​≤M. Then, the rectangle is composed of all the cell (x, y) where x1≤x≤x2x_1 \leq x \leq x_2x1​≤x≤x2​ and y1≤y≤y2y_1 \leq y \leq y2y1​≤y≤y2. If all of the cell in the rectangle is "1"\texttt{"1"}"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.

输入描述:

The first line of input contains two space-separated integers N and M.

Following N lines each contains M characters cijc_{ij}cij​.

1≤N,M≤10001 \leq N, M \leq 10001≤N,M≤1000

N×M≥2N \times M \geq 2N×M≥2

cij∈"01"c_{ij} \in \texttt{"01"}cij​∈"01"

输出描述:

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

输入

1 2
01

输出

0

输入

1 3
101

输出

1

题意：

给出只包含0和1的矩形长和宽，计算出第二大全1矩阵的面积。

思路：

a.先简化成求最大全1矩阵面积：

1、对01矩阵的每一行求列（行）前缀和

2、在每一行中找每个元素的左右边界

例如：

输入矩阵：

1110

1110

0011

0011

第一步：（求列前缀和）

1110

2220

0031

0042

第二步：找左右边界（取第四行 0042 为例）

为便于观察，先画出分布图

对于图中每一个列元素，以其为高的最大矩形边长分别是：

l[1]=0,r[1]=5  S=0*(5-0-1)=0

l[2]=0,r[2]=5  S=0*(5-0-1)=0

l[3]=2,r[3]=4  S=4*(4-2-1)=4

l[1]=2,r[1]=5  S=2*(5-2-1)=4

所以最大面积为4

b.根据左右区间求出了每一行的最大面积后

可以知道 第二大矩形的面积=所有矩形中第二大的面积/最大面积的长或宽-1后相乘

具体实现：

对每一行中的列元素查找左右区间，要利用单调栈 具体讲解：单调栈

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e3;
int n,m;
int high[MAX+5][MAX+5],Stack[MAX+5],l[MAX+5],r[MAX+5];
char a[MAX+5][MAX+5];
int cnt;
int num[5*MAX*MAX+5];
bool flag[MAX+5][MAX+5];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        high[i][0]=-1,high[i][m+1]=-1;
        scanf("%s",a[i]+1);
        for(int j=1;j<=m;j++){
            if(a[i][j]=='0'){
                high[i][j]=0;
                continue;
            }
            if(i==0)
                high[i][j]=a[i][j]-'0';
            else
                high[i][j]=high[i-1][j]+(a[i][j]-'0');
        }
    }
    for(int i=0;i<n;i++){
        int top=0;
        Stack[top]=0;
        for(int j=1;j<=m;j++){
            while(high[i][Stack[top]]>=high[i][j]) top--;
            l[j]=Stack[top];
            Stack[++top]=j;
        }
        top=0;
        Stack[top]=m+1;
        for(int j=m;j>=1;j--){
            while(high[i][Stack[top]]>=high[i][j]) top--;
            r[j]=Stack[top];
            Stack[++top]=j;
        }
        memset(flag,false,sizeof(flag));
        for(int j=1;j<=m;j++){
            int x=r[j]-l[j]-1,y=high[i][j];
            if(!y) continue;
            if(!flag[l[j]][r[j]]){ ///标记一行中出现过的左右区间，否则同一个矩形面积会多次计算
                num[cnt++]=x*y;
                num[cnt++]=(x-1)*y;
                num[cnt++]=x*(y-1);
                flag[l[j]][r[j]]=true;
            }
        }
    }
    sort(num,num+cnt);
    printf("%d\n",num[cnt-2]);
    return 0;
}
/*
1 7
2145133
0 0 2 3 0 5 5
2 8 5 5 8 8 8
*/