POJ-3494 Largest Submatrix of All 1’s （单调栈）

Largest Submatrix of All 1’s

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 8551		Accepted: 3089
Case Time Limit: 2000MS

Description

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.

Input

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 nnumbers. The input ends once EOF is met.

Output

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.

Sample Input

2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0

Sample Output

0
4

题意：

求内部全部都是1的子矩阵最大面积。

思路：

第一时间想到了一个n的3次方的dp，但是这样肯定超时。

类似的题 ：https://www.cnblogs.com/ZGQblogs/p/10664506.html

如果不是在学单调栈，我觉得我一定是想不到的。

首先就是维护每一行的每一个位置，如果以当前行为底，上面连续的1有多少个。

然后就是这个题：http://poj.org/problem?id=2559

代码：

加了读入挂才过～～～

 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<stack>
 #include<queue>
 #include<map>
 #include<set>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<ctime>
 #define fuck(x) cout<<#x<<" = "<<x<<endl;
 #define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
 #define ls (t<<1)
 #define rs ((t<<1)+1)
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 const int maxn = ;
 const int maxm = ;
 const int inf = 2.1e9;
 const ll Inf = ;
 const int mod = ;
 const double eps = 1e-;
 const double pi = acos(-);

 int l[maxn],r[maxn];
 struct node{
     int num,pos;
 };
 int mp[maxn][maxn];
 stack<node>st;

 int solve(int *num,int n){
     for(int i=;i<=n;i++){
         l[i]=r[i]=i;
     }
     num[]=num[n+]=-;
     for(int i=;i<=n+;i++){
         while(!st.empty()&&st.top().num>num[i]){
             r[st.top().pos]=i-;
             st.pop();
         }
         st.push(node{num[i],i});
     }
     while(!st.empty()){st.pop();}

     for(int i=n;i>=;i--){
         while(!st.empty()&&st.top().num>num[i]){
             l[st.top().pos]=i+;
             st.pop();
         }
         st.push(node{num[i],i});
     }
     ll ans=;
     for(int i=;i<=n;i++){
         ans=max(ans,1ll*num[i]*(r[i]-l[i]+));
     }
     while(!st.empty()){st.pop();}
     return ans;
 }

 char buf[maxn], *ps = buf, *pe = buf+;
 inline void rnext(){
     if(++ps == pe)
         pe = (ps = buf)+fread(buf,,sizeof(buf),stdin);
 }
 template <class T>
 inline bool in(T &ans)
 {
     ans = ;
     T f = ;
     if(ps == pe) return false;
     do{
         rnext();
         if('-' == *ps) f = -;
     }while(!isdigit(*ps) && ps != pe);
     if(ps == pe) return false;
     do
     {
         ans = (ans<<)+(ans<<)+*ps-;
         rnext();
     }while(isdigit(*ps) && ps != pe);
     ans *= f;
     return true;
 }

 int main()
 {
 //    freopen("in.txt","r",stdin);
     int n,m;
     while(true){
         in(n);in(m);
         if(!m||!n){break;}
         int ans=;
         for(int i=;i<=n;i++){
             for(int j=;j<=m;j++){
                 in(mp[i][j]);
 //                cout<<mp[i][j]<<endl;
                 if(mp[i][j]==){mp[i][j]+=mp[i-][j];}
             }
             ans=max(ans,solve(mp[i],m));
         }
         printf("%d\n",ans);
     }
     return ;
 }