題目鏈接:https://ac.nowcoder.com/acm/contest/882/H
題目描述
Given a N×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 x1,y1,x2,y2
where 1≤x1≤x2≤N and 1≤y1≤y2≤M. Then, the rectangle is composed of all the cell (x, y) where x1≤x≤x2 and y1≤y≤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.
輸入描述:
The first line of input contains two space-separated integers N and M.
Following N lines each contains M characters cij
. 1≤N,M≤1000
N×M≥2
cij∈"01"
輸出描述:
Output one line containing an integer representing the answer. If there are less than 2 rectangles containning all "1"
, output "0"
.
示例1
輸入
1 2 01
輸出
0
示例2
輸入
1 3 101
輸出
1
題面大意:
給出一個由0和1組成的一個矩陣,求出出這個矩陣中全部由1組成的第二大的子矩陣
需要使用單調棧
對矩陣進行處理, 轉化成一個直方圖,然後套用單調棧即可
代碼:
#include<iostream>
#include<stdio.h>
#include<stack>
using namespace std;
int max1=0,max2=0;
int a[1010][1010];
char b[1010];
void check(int n){
if(n>max1){
int t=max1;
max1=n;
max2=t;
}
else if(n>max2){
max2=n;
}
}
void Area(int x,int y){
check(x*y);
check((x-1)*y);
check(x*(y-1));
}
int main(){
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++){
cin>>b;
for(int j=1;j<=m;j++){
a[i][j]=b[j-1]-'0';
if(a[i][j]==1){
a[i][j]+=a[i-1][j];
}
}
}
for(int i=1;i<=n;i++){
a[i][0]=-2;
a[i][m+1]=-1;
stack<int> q;
q.push(0);
for(int j=1;j<=m+1;j++){
while(a[i][q.top()]>a[i][j]){
int index=q.top();
q.pop();
Area(j-q.top()-1,a[i][index]);
}
q.push(j);
}
}
cout<<max2<<endl;
}