矩阵连乘动态规划C语言代码

以下是矩阵连乘动态规划的 C 语言代码示例：

#include <stdio.h>

#define MAX_SIZE 50

void matrixChainOrder(int p[], int n, int m[MAX_SIZE][MAX_SIZE], int s[MAX_SIZE][MAX_SIZE]) {
    int i, j, k, len, q;
    for (i = 1; i <= n; i++) {
        m[i][i] = 0;
    }
    for (len = 2; len <= n; len++) {
        for (i = 1; i <= n - len + 1; i++) {
            j = i + len - 1;
            m[i][j] = 99999999;
            for (k = i; k <= j - 1; k++) {
                q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (q < m[i][j]) {
                    m[i][j] = q;
                    s[i][j] = k;
                }
            }
        }
    }
}

void printOptimalParens(int s[MAX_SIZE][MAX_SIZE], int i, int j) {
    if (i == j) {
        printf("A%d", i);
    } else {
        printf("(");
        printOptimalParens(s, i, s[i][j]);
        printOptimalParens(s, s[i][j] + 1, j);
        printf(")");
    }
}

int main() {
    int n, p[MAX_SIZE], m[MAX_SIZE][MAX_SIZE], s[MAX_SIZE][MAX_SIZE];
    printf("Enter number of matrices: ");
    scanf("%d", &n);
    printf("Enter dimensions of matrices:\n");
    for (int i = 0; i <= n; i++) {
        scanf("%d", &p[i]);
    }
    matrixChainOrder(p, n, m, s);
    printf("Optimal Parenthesization: ");
    printOptimalParens(s, 1, n);
    printf("\nMinimum number of multiplications: %d\n", m[1][n]);
    return 0;
}

希望这个例子能帮到您！