分支限界法求解0-1背包问题,规定物品数量为10，背包容量为50，输入为20个数，前十个为物品重量，后十个数为物品价值,键盘输入C语言代码

以下是使用分支限界法求解0-1背包问题的C语言代码：

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

#define N 10 // 物品数量
#define C 50 // 背包容量

typedef struct {
    int weight; // 物品重量
    int value; // 物品价值
    float bound; // 松弛上界
    int level; // 结点所在层数
} Node;

int max_value = 0; // 最大价值
int max_weight[N+1] = {0}; // 最大价值对应的物品重量

void knapsack(int *w, int *v) {
    Node queue[100]; // 结点队列
    int front = 0, rear = 0; // 队头和队尾指针
    Node u, v1;

    // 初始化根结点
    u.level = 0;
    u.weight = 0;
    u.value = 0;
    u.bound = 0;
    queue[rear++] = u;

    while (front < rear) {
        // 取出队头结点
        u = queue[front++];
        if (u.level == N) { // 叶结点
            if (u.value > max_value) {
                max_value = u.value;
                for (int i = 0; i < N; i++) {
                    max_weight[i] = 0;
                }
                for (int i = N-1; i >= 0; i--) {
                    if (u.weight >= w[i]) {
                        max_weight[i] = 1;
                        u.weight -= w[i];
                    }
                    else {
                        max_weight[i] = u.weight / w[i];
                        break;
                    }
                }
            }
            continue;
        }

        // 计算左儿子结点的松弛上界
        v1.level = u.level + 1;
        v1.weight = u.weight + w[v1.level - 1];
        v1.value = u.value + v[v1.level - 1];
        if (v1.weight <= C) {
            v1.bound = v1.value;
            for (int i = v1.level; i < N; i++) {
                if (v1.weight + w[i] <= C) {
                    v1.weight += w[i];
                    v1.value += v[i];
                }
                else {
                    v1.bound += (C - v1.weight) * 1.0 / w[i] * v[i];
                    break;
                }
            }
            if (v1.bound > max_value) { // 产生新结点
                queue[rear++] = v1;
            }
        }

        // 计算右儿子结点的松弛上界
        v1.level = u.level + 1;
        v1.weight = u.weight;
        v1.value = u.value;
        v1.bound = v1.value;
        for (int i = v1.level; i < N; i++) {
            if (v1.weight + w[i] <= C) {
                v1.weight += w[i];
                v1.value += v[i];
            }
            else {
                v1.bound += (C - v1.weight) * 1.0 / w[i] * v[i];
                break;
            }
        }
        if (v1.bound > max_value) { // 产生新结点
            queue[rear++] = v1;
        }
    }
}

int main() {
    int w[N], v[N];
    printf("请输入%d个数，前%d个为物品重量，后%d个为物品价值：\n", N*2, N, N);
    for (int i = 0; i < N; i++) {
        scanf("%d", &w[i]);
    }
    for (int i = 0; i < N; i++) {
        scanf("%d", &v[i]);
    }

    knapsack(w, v);

    printf("最大价值：%d\n", max_value);
    printf("选择的物品重量为：");
    for (int i = 0; i < N; i++) {
        printf("%d ", max_weight[i] * w[i]);
    }
    printf("\n");

    return 0;
}