Maximum sum rectangle in a 2D matrix | DP-27 in C++

ในบทช่วยสอนนี้ เราจะพูดถึงโปรแกรมหาสี่เหลี่ยมผืนผ้าที่มีผลรวมสูงสุดในเมทริกซ์ 2 มิติ

เราจะได้รับเมทริกซ์ งานคือหาเซ็บเมทริกซ์ที่มีผลรวมขององค์ประกอบสูงสุด

ตัวอย่างโค้ด

#include <bits/stdc++.h>
using namespace std;

#define ROW 4
#define COL 5

int kadane(int* arr, int* start, int* finish, int n) {
    int sum = 0, maxSum = INT_MIN, i;
    *finish = -1;
    int local_start = 0;
    for (i = 0; i < n; ++i) {
        sum += arr[i];
        if (sum < 0) {
            sum = 0;
            local_start = i + 1;
        } else if (sum > maxSum) {
            maxSum = sum;
            *start = local_start;
            *finish = i;
        }
    }
    if (*finish != -1) return maxSum;
    maxSum = arr[0];
    *start = *finish = 0;
    for (i = 1; i < n; i++) {
        if (arr[i] > maxSum) {
            maxSum = arr[i];
            *start = *finish = i;
        }
    }
    return maxSum;
}

void findMaxSum(int M[][COL]) {
    int maxSum = INT_MIN, finalLeft, finalRight, finalTop, finalBottom;
    int left, right, i;
    int temp[ROW], sum, start, finish;
    for (left = 0; left < COL; ++left) {
        memset(temp, 0, sizeof(temp));
        for (right = left; right < COL; ++right) {
            for (i = 0; i < ROW; ++i)
                temp[i] += M[i][right];
            sum = kadane(temp, &start, &finish, ROW);
            if (sum > maxSum) {
                maxSum = sum;
                finalLeft = left;
                finalRight = right;
                finalTop = start;
                finalBottom = finish;
            }
        }
    }
    cout << "(Top, Left) (" << finalTop << ", " << finalLeft << ")" << endl;
    cout << "(Bottom, Right) (" << finalBottom << ", " << finalRight << ")" << endl;
    cout << "Max sum is: " << maxSum << endl;
}

int main() {
    int M[ROW][COL] = {
        {1, 2, -1, -4, -20},
        {-8, -3, 4, 2, 1},
        {3, 8, 10, 1, 3},
        {-4, -1, 1, 7, -6}
    };
    findMaxSum(M);
    return 0;
}

เอาต์พุต

(Top, Left) (1, 1) (Bottom, Right) (3, 3) Max sum is: 29