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.
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.
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.
2 2 0 0 0 0 4 4 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0Sample Output
0 4
单调栈，每一行记录个列的高度 然后每一行进行单调栈，用输入输出流还是超时得用scanf。。或者 用scanf就不错。。
#include <iostream> #include <cstdio> #include <cstring> #include <map> #include <queue> #include <cmath> using namespace std; int n,m,d; int mp[2002][2002]; int l[2002]={1},r[2002],stack[2001],top,maxi,sum; int main() { //ios::sync_with_stdio(false); //cin.tie(0); while(scanf("%d%d",&n,&m)!=EOF) { maxi=0; for(int i=1;i<=n;i++) { top=1; for(int j=1;j<=m;j++) { scanf("%d",&d); //mp[i][j]=mp[i][j-1]+d; 不对的 中间如果有0 应该中断 mp[i][j]=(d==0?0:(mp[i-1][j]+d)); l[j]=r[j]=j; } for(int j=1;j<=m+1;j++) { while(top>1&&mp[i][j]<=mp[i][stack[top-1]]) { r[stack[top-1]]=j-1; sum=(r[stack[top-1]]-l[stack[top-1]]+1)*mp[i][stack[top-1]]; if(sum>maxi)maxi=sum; top--; } l[j]=stack[top-1]+1; stack[top++]=j; } } printf("%d\n",maxi); } } /* 4 4 1 0 1 1 0 1 1 1 0 0 1 1 0 1 1 1 */
原文：http://www.cnblogs.com/8023spz/p/7239162.html