C ++程序中最大子矩阵面积为1的计数比0的计数大
在这个问题中,我们得到一个大小为nXn的二维矩阵,其中包含二进制数(0/1)。我们的任务是创建一个程序,以找到最大子矩阵区域,该区域的计数为1的点数大于0的数。
让我们举个例子来了解这个问题,
输入项
bin[N][N] = {{0, 1, 0, 0},
{1, 1, 0, 0},
{1, 0, 1, 1},
{0, 1, 0, 1}
}
输出结果
9
说明
submatrix :bin[1][0], bin[1][1], bin[1][2]
bin[2][0], bin[2][1], bin[2][2]
bin[3][0], bin[3][1], bin[3][2]
is the largest subarray with more 1’s one more than 0’s.
Number of 0’s = 4
Number of 1’s = 5
解决方法
一种简单的方法是从矩阵中找到所有可能的子矩阵,然后从所有矩阵中返回最大面积。
这种方法易于思考和实现,但是要花费大量时间,并且需要嵌套循环,从而使时间复杂度为O(n ^ 4)。因此,让我们讨论另一种更有效的方法。
这里的想法是将列固定在矩阵的左右两边,然后找到最大的子数组,该子数组的数字为0的数字大于1的数字。我们将计算每一行的总和,然后将其累加。查找计数为1的最大区域多于0的数目。
示例
该程序说明了我们解决方案的工作原理,
#include <bits/stdc++.h>using namespace std;
#define SIZE 10
int lenOfLongSubarr(int row[], int n, int& startInd, int& finishInd){
unordered_map<int, int> subArr;
int sumVal = 0, maxSubArrLen = 0;
for (int i = 0; i < n; i++) {
sumVal += row[i];
if (sumVal == 1) {
startInd = 0;
finishInd = i;
maxSubArrLen = i + 1;
}
else if (subArr.find(sumVal) == subArr.end())
subArr[sumVal] = i;
if (subArr.find(sumVal − 1) != subArr.end()) {
int currLen = (i − subArr[sumVal − 1]);
if (maxSubArrLen < currLen)
startInd = subArr[sumVal − 1] + 1;
finishInd = i;
maxSubArrLen = currLen;
}
}
return maxSubArrLen;
}
int largestSubmatrix(int bin[SIZE][SIZE], int n){
int rows[n], maxSubMatArea = 0, currArea, longLen, startInd,
finishInd;
for (int left = 0; left < n; left++) {
memset(rows, 0, sizeof(rows));
for (int right = left; right < n; right++) {
for (int i = 0; i < n; ++i){
if(bin[i][right] == 0)
rows[i] −= 1;
else
rows[i] += 1;
}
longLen = lenOfLongSubarr(rows, n, startInd, finishInd);
currArea = (finishInd − startInd + 1) * (right − left + 1);
if ((longLen != 0) && (maxSubMatArea < currArea)) {
maxSubMatArea = currArea;
}
}
}
return maxSubMatArea;
}
int main(){
int bin[SIZE][SIZE] = {
{ 1, 0, 0, 1 },
{ 0, 1, 1, 1 },
{ 1, 0, 0, 0 },
{ 0, 1, 0, 1 }
};
int n = 4;
cout<<"The maximum sub−matrix area having count of 1’s one more
than count of 0’s is "<<largestSubmatrix(bin, n);
return 0;
}
输出结果
The maximum sub-matrix area having count of 1’s one more than count of0’s is 9
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