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題目連結:https://leetcode.com/problems/maximal-square/
題意: 給出一個隻包含0,1的二維矩陣,要求找到一個全為1的正方形的子矩陣,并輸出子矩陣的面積
思路: 詳細解釋請參考Maximal Rectangle,在這裡隻需要增加一個判斷即可,那就是r==c
class Solution
{
public:
int maximalSquare(vector<vector<char>>& matrix)
{
int n = matrix.size();
if(n==0) return 0;
int m = matrix[0].size();
int i,j,c;
vector<vector<int> > dp,a;
dp.resize(n+1),a.resize(n+1);
for(i = 0; i<=n; i++)
{
dp[i].resize(m+1);
a[i].resize(m+1);
}
for(i = 0; i<n; i++)
{
for(j = 0; j<m; j++)
{
a[i+1][j+1] = matrix[i][j]-'0';
}
}
int sum = 0;
for(i = 1; i<=m; i++)
{
sum+=a[1][i];
dp[1][i] = sum;
}
for(i = 2; i<=n; i++)
{
sum = 0;
for(j = 1; j<=m; j++)
{
sum+=a[i][j];
dp[i][j]=dp[i-1][j]+sum;
}
}
int maxn = 0;
for(i = n; i>0; i--)
{
for(j = m; j>0&&maxn<i*j; j--)
{
if(a[i][j])
{
int r = 1,c = 1;
while(j-c>=0)
{
if(dp[i][j]-dp[i][j-c]-dp[i-r][j]+dp[i-r][j-c]==r*c)
{
if(r==c)
maxn = max(maxn,r*c);
c++;
}
else
break;
}
while(i-r>=0&&c>0)
{
if(dp[i][j]-dp[i][j-c]-dp[i-r][j]+dp[i-r][j-c]==r*c)
{
if(r==c)
maxn = max(maxn,r*c);
r++;
}
else
c--;
}
}
}
}
return maxn;
}
};