2019牛客多校第八场——A. All-one Matrices【单调栈/思维】【不可扩大的全1子矩阵个数】
链接：https://ac.nowcoder.com/acm/contest/888/A
题目描述
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.
输入描述:
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
输出描述:
Print a non-negative integer, denoting the answer.输入
3 4 0111 1110 0101输出
5说明
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).
题意：
给你一个01矩阵，求出所有不可扩大的全1矩阵的个数；
分析：
对于每一个[i,j]，记录从它开始向上连续的1的个数num；
枚举每一行作为矩阵底边所在行，从前向后枚举每一列，维护一个num单调上升的栈，对于栈中每一个num值，还要维护一个该num对应位置向左最远能扩展到的位置pos；
元素出栈时，设该元素是（num，pos），那么可以得到一个全1矩阵（i-num+1，pos）->（i，j）；
此时该矩阵最左为pos已经记录（不可向左扩展），向上最多为num已经记录（不可向上扩展），因为大于当前num才出栈（不可向右扩展），所以只要判定能否向下扩展，如果不能，那么这就是一个答案矩阵。
#include
using namespace std;
const int maxn=3007;
char s[maxn][maxn];
int num[maxn][maxn];
stack< pair > S;
int main()
{
int n,m;scanf("%d%d",&n,&m);getchar();
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++) scanf("%c",&s[i][j]);
getchar();
}
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
num[i][j]=(s[i][j]=='1')?num[i-1][j]+1:0;
int ans=0;
for (int i=1;i<=n;i++)
{
int tmp=-1;
while (!S.empty()) S.pop();
for (int j=1;j<=m+1;j++)
{
int pos=j;
while (!S.empty() && S.top().first>num[i][j])
{
if(S.top().second<=tmp) ans++;//下一行中[pos,j]中有0存在 不可向下扩展
pos=S.top().second;
S.pop();
}
if(!num[i+1][j]) tmp=j;//记录下一行中截至当前最右边0的位置
if(num[i][j] && (S.empty() || S.top().first