假设一个0/1背包问题是，n=4，重量为w=（4，7，5，3），价值为v=（40，42，25，12），背包限重为W=10，解向量为x=（x1，x2，x3，x4）。请采用课本第6章中的优先队列式分枝限界法求解该问题，C语言

以下是使用C语言实现优先队列式分枝限界法解决0/1背包问题的示例代码：

#include <stdio.h>
#include <stdlib.h>

#define MAX_N 100 // 最大物品数量
#define MAX_W 1000 // 最大背包容量

struct node {
    int level; // 当前搜索层数
    int profit; // 当前价值
    int weight; // 当前重量
    int bound; // 上界
};

struct queue_node {
    struct node *data; // 存储结点指针
    struct queue_node *next; // 指向下一个结点的指针
};

struct queue {
    struct queue_node *front; // 队首指针
    struct queue_node *rear; // 队尾指针
};

// 初始化队列
void init_queue(struct queue *q) {
    q->front = q->rear = NULL;
}

// 判断队列是否为空
int is_queue_empty(struct queue *q) {
    return q->front == NULL;
}

// 入队
void enqueue(struct queue *q, struct node *data) {
    struct queue_node *new_node = (struct queue_node *) malloc(sizeof(struct queue_node));
    new_node->data = data;
    new_node->next = NULL;

    if (is_queue_empty(q)) {
        q->front = q->rear = new_node;
    } else {
        q->rear->next = new_node;
        q->rear = new_node;
    }
}

// 出队
struct node *dequeue(struct queue *q) {
    if (is_queue_empty(q)) {
        return NULL;
    }

    struct queue_node *front_node = q->front;
    struct node *data = front_node->data;

    q->front = front_node->next;
    free(front_node);

    if (q->front == NULL) {
        q->rear = NULL;
    }

    return data;
}

// 计算结点的上界
int calc_bound(int level, int profit, int weight, int n, int *w, int *v, int W) {
    int bound = profit;
    int j = level + 1;
    int total_weight = weight;

    while (j < n && total_weight + w[j] <= W) {
        bound += v[j];
        total_weight += w[j];
        j++;
    }

    if (j < n) {
        bound += (W - total_weight) * v[j] / w[j];
    }

    return bound;
}

// 初始化根结点
void init_root_node(struct node *root, int n, int *w, int *v, int W) {
    root->level = -1;
    root->profit = 0;
    root->weight = 0;
    root->bound = calc_bound(-1, 0, 0, n, w, v, W);
}

// 初始化子结点
void init_child_node(struct node *child, struct node *parent, int i, int n, int *w, int *v, int W) {
    child->level = parent->level + 1;
    child->profit = parent->profit + v[i];
    child->weight = parent->weight + w[i];
    child->bound = calc_bound(child->level, child->profit, child->weight, n, w, v, W);
}

// 比较函数，用于优先队列中的结点排序
int compare_node(const void *a, const void *b) {
    struct node *node_a = *((struct node **) a);
    struct node *node_b = *((struct node **) b);

    return node_b->bound - node_a->bound;
}

// 优先队列式分枝限界法求解0/1背包问题
int knapsack(int n, int *w, int *v, int W, int *x) {
    struct node root, *p, *left_child, *right_child, *max_node;
    int max_profit = 0;
    struct queue q;

    init_root_node(&root, n, w, v, W);
    init_queue(&q);
    enqueue(&q, &root);

    while (!is_queue_empty(&q)) {
        p = dequeue(&q);
        if (p->bound > max_profit) {
            left_child = (struct node *) malloc(sizeof(struct node));
            right_child = (struct node *) malloc(sizeof(struct node));
            init_child_node(left_child, p, p->level + 1, n, w, v, W);
            init_child_node(right_child, p, p->level + 1, n, w, v, W);

            if (left_child->weight <= W && left_child->profit > max_profit) {
                max_profit = left_child->profit;
                max_node = left_child;
            }

            if (left_child->bound > max_profit) {
                enqueue(&q, left_child);
            } else {
                free(left_child);
            }

            if (right_child->weight <= W && right_child->profit > max_profit) {
                max_profit = right_child->profit;
                max_node = right_child;
            }

            if (right_child->bound > max_profit) {
                enqueue(&q, right_child);
            } else {
                free(right_child);
            }
        }

        free(p);
    }

    for (int i = n - 1; i >= 0; i--) {
        if (max_node->weight >= w[i]) {
            x[i] = 1;
            max_node->weight -= w[i];
        } else {
            x[i] = 0;
        }
    }

    return max_profit;
}

int main() {
    int n = 4;
    int w[MAX_N] = {4, 7, 5, 3};
    int v[MAX_N] = {40, 42, 25, 12};
    int W = 10;
    int x[MAX_N] = {0};

    int max_profit = knapsack(n, w, v, W, x);

    printf("最大价值为：%d\n", max_profit);
    printf("解向量为：(");

    for (int i = 0; i < n; i++) {
        printf("%d", x[i]);
        if (i < n - 1) {
            printf(", ");
        }
    }

    printf(")\n");

    return 0;
}

该代码实现了优先队列式分枝限界法求解0/1背包问题，并输出了最大价值和解向量。