标签：end view cli nts += ide ace font one
描述
Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.
输入
The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m lines each with n numbers. The input ends once EOF is met.
输出
For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.
样例输入
2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0
样例输出
0
4
高：是1累加上前一行的，0的话就是0；
长：把每一行作为长，算mp【i】【j】的时候分别向左向右找第一个比它小的数，所以用单调递减栈。
这有个讲单调栈和单调队列的讲的超级好，我觉得（https://www.cnblogs.com/tham/p/8038828.html ）
#include<bits/stdc++.h> using namespace std; int mp[2222][2222]; int l[2222],r[2222]; int main() { int n,m; while(~scanf("%d%d",&n,&m)) { for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++) { scanf("%d",&mp[i][j]); if(mp[i][j]) mp[i][j]+=mp[i-1][j]; } } for(int i=1; i<=n; i++) { mp[i][0]=-1; mp[i][m+1]=-1; } int mx=0; stack<int>s; while(!s.empty())s.pop(); for(int i=1; i<=n; i++) { while(!s.empty())s.pop(); s.push(0); for(int j=1; j<=m; j++)//在前面找第一个比自己小的 { while(!s.empty()&&mp[i][j]<=mp[i][s.top()])s.pop();//当前元素小于等于栈顶 出栈 l[j]=s.top();//记录下标 s.push(j); } while(!s.empty())s.pop(); s.push(m+1); for(int j=m; j>=1; j--)//在后面找第一个比自己小的 { while(!s.empty()&&mp[i][j]<=mp[i][s.top()])s.pop(); r[j]=s.top(); s.push(j); } int sum=0; for(int j=1; j<=m; j++) { sum=mp[i][j]*(r[j]-l[j]-1); mx=max(mx,sum); } } cout<<mx<<endl; } return 0; }
2906: Largest Submatrix of All 1’s
标签：end view cli nts += ide ace font one
原文地址：https://www.cnblogs.com/ydw--/p/10971440.html
