鏈接：https://ac.nowcoder.com/acm/contest/888/A
來源：牛客網
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 524288K，其他語言1048576K
64bit IO Format: %lld
題目描述
Gromah and LZR entered the great tomb, the first thing they see is a matrix of size n×mn\times mn×m, and the elements in the matrix are all 00_{}0 or 11_{}1.
LZR finds a note board saying "An all-one matrix is defined as the matrix whose elements are all 11_{}1, you should determine the number of all-one submatrices of the given matrix that are not completely included by any other all-one submatrices".
Meanwhile, Gromah also finds a password lock, obviously the password should be the number mentioned in the note board!
Please help them determine the password and enter the next level.
輸入描述:
The first line contains two positive integers n,mn,m_{}n,m, denoting the size of given matrix.
Following nn_{}n lines each contains a string with length mm_{}m, whose elements are all 00_{}0 or 11_{}1, denoting the given matrix.
1≤n,m≤30001\le n,m \le 30001≤n,m≤3000
輸出描述:
Print a non-negative integer, denoting the answer.
示例1
輸入
3 4 0111 1110 0101
輸出
5
說明
The 5 matrices are (1,2)−(1,4), (1,2)−(2,3), (1,2)−(3,2), (2,1)−(2,3), (3,4)−(3,4)(1,2)-(1,4), \; (1,2)-(2,3), \; (1,2)-(3,2), \; (2,1)-(2,3), \; (3,4)-(3,4)_{}(1,2)−(1,4),(1,2)−(2,3),(1,2)−(3,2),(2,1)−(2,3),(3,4)−(3,4).
題目大意：給出了一個01矩陣，讓求有多少個全一矩陣不被其他全1矩陣覆蓋。
解題思路：預處理出每一個點向上連續的1是多少即h[i][j]，即確定上邊界。
我們依次枚舉下邊界第i行，對於當前點i,j h[i][j] 爲上邊界，然後用單調棧處理出 以上邊界爲最小值的左右邊界L，R。
然後只需要判斷一下，當前的下邊界能不能向下延伸就可以了。注意對於第i行同一個子矩陣可能被多次計算（高度相同），
所以我們維護一個高度遞減的單調棧就可以了。
#include<bits/stdc++.h>
using namespace std;
const int N = 3005;
typedef long long ll;
const int mmax = 1e5+5;
char s[N][N];
int sum[N][N],L[N],R[N];
int h[N][N];
stack<int>st;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)scanf("%s",s[i]+1);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if(s[i][j]-'0')h[i][j] = h[i-1][j]+1;
sum[i][j] = sum[i][j-1] + s[i][j] - '0';
}
}
int ans=0;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
while(!st.empty() && h[i][j] <= h[i][st.top()]) st.pop();
if(st.empty()) L[j] = 1;
else L[j] = st.top()+1;
st.push(j);
}
while(!st.empty())st.pop();
for(int j=m; j>=1; j--)
{
while(!st.empty() && h[i][j] <= h[i][st.top()]) st.pop();
if(st.empty()) R[j] = m;
else R[j] = st.top()-1;
st.push(j);
}
while(!st.empty())st.pop();
for(int j=m; j>=1; j--)
{
if(h[i][j]==0)
{
while(!st.empty())st.pop();
continue;
}
while(!st.empty() && h[i][j] < h[i][st.top()]) st.pop();
if(st.empty() || h[i][j] != h[i][st.top()]) //避免高度相同的多次計算。
{
int l = L[j];
int r = R[j];
if(sum[i+1][r]-sum[i+1][l-1] != r - l + 1)ans++;
}
st.push(j);
}
while(!st.empty())st.pop();
}
cout<<ans<<endl;
}