输油管道问题分治法c语言代码简单

以下是一个简单的输油管道问题的分治法 C 语言代码：

#include <stdio.h>

// 求解输油管道问题
int pipeline(int *profits, int start, int end) {
    if (start == end) {
        return profits[start];
    }
    int mid = (start + end) / 2;
    int left = pipeline(profits, start, mid);
    int right = pipeline(profits, mid + 1, end);
    int cross = profits[mid] + profits[mid + 1];
    int sum = cross;
    for (int i = mid - 1, temp = cross; i >= start; i--) {
        temp += profits[i];
        if (temp > sum) {
            sum = temp;
        }
    }
    for (int i = mid + 2, temp = cross; i <= end; i++) {
        temp += profits[i];
        if (temp > sum) {
            sum = temp;
        }
    }
    if (left > right && left > sum) {
        return left;
    } else if (right > left && right > sum) {
        return right;
    } else {
        return sum;
    }
}

int main() {
    int profits[] = {2, -1, 3, 4, -5, 6, -7, 8};
    int n = sizeof(profits) / sizeof(int);
    int max_profit = pipeline(profits, 0, n - 1);
    printf("The maximum profit is %d\n", max_profit);
    return 0;
}

该代码的主要思路是利用分治法将输油管道问题分成两个子问题，分别求出左边子数组的最大收益、右边子数组的最大收益和跨越中间的最大收益，最后比较三者的大小，取最大值作为整个问题的解。