先进先出队列式分支限界法求解0-1背包问题用c++完整代码

以下是使用先进先出队列式分支限界法求解0-1背包问题的完整C代码：

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

#define MAX_SIZE 100

//物品结构体
typedef struct _item{
    int weight; //重量
    int value;  //价值
    int bound;  //价值上界
}Item;

//结点结构体
typedef struct _node{
    int level;        //决策树层数
    int profit;       //当前已获得的价值
    int weight;       //当前已占用的重量
    int bound;        //价值上界
    int select[MAX_SIZE];  //选择情况
}Node;

//队列结构体
typedef struct _queue{
    Node elem[MAX_SIZE];  //队列元素
    int front;        //队头指针
    int rear;         //队尾指针
}Queue;

//初始化队列
void initQueue(Queue *q){
    q->front = q->rear = 0;
}

//判断队列是否为空
int isEmpty(Queue *q){
    if(q->front == q->rear)
        return 1;
    else
        return 0;
}

//进队列
void enqueue(Queue *q, Node n){
    if((q->rear+1)%MAX_SIZE == q->front){
        printf("Queue is full!\n");
        exit(1);
    }
    q->elem[q->rear] = n;
    q->rear = (q->rear+1)%MAX_SIZE;
}

//出队列
Node dequeue(Queue *q){
    if(isEmpty(q)){
        printf("Queue is empty!\n");
        exit(1);
    }
    Node n = q->elem[q->front];
    q->front = (q->front+1)%MAX_SIZE;
    return n;
}

//计算结点的价值上界
int bound(Node n, int nItems, Item items[]){
    int j, k;
    int totalWeight;
    int boundValue;

    //剩余物品全部装入背包
    if(n.weight >= items[n.level].weight){
        boundValue = n.profit;
        totalWeight = n.weight;
        for(j=n.level+1; j<nItems; j++){
            if(totalWeight+items[j].weight <= MAX_SIZE){
                totalWeight += items[j].weight;
                boundValue += items[j].value;
            }else{
                k = MAX_SIZE-totalWeight;
                boundValue += (int)(k*(items[j].value/items[j].weight));
                break;
            }
        }
    }
    //剩余物品不能全部装入背包
    else{
        boundValue = n.profit+(int)((MAX_SIZE-n.weight)*(items[n.level].value/items[n.level].weight));
        totalWeight = MAX_SIZE;
    }

    return boundValue;
}

//先进先出队列式分支限界法
int knapsack(int nItems, Item items[], int capacity, int *solution){
    Queue q;
    Node u, v;
    int i;

    initQueue(&q);

    //初始化根结点
    u.level = -1;
    u.profit = 0;
    u.weight = 0;

    //计算根结点的价值上界
    u.bound = bound(u, nItems, items);

    enqueue(&q, u);

    int maxProfit = 0;

    while(!isEmpty(&q)){
        u = dequeue(&q);

        //如果结点的价值上界小于当前最优解，则剪枝
        if(u.bound <= maxProfit)
            continue;

        //扩展结点
        if(u.level < nItems-1){
            //不选当前物品
            v.level = u.level+1;
            v.weight = u.weight;
            v.profit = u.profit;
            v.bound = bound(v, nItems, items);
            for(i=0; i<=u.level; i++){
                v.select[i] = u.select[i];
            }
            v.select[v.level] = 0;
            enqueue(&q, v);

            //选当前物品
            v.level = u.level+1;
            v.weight = u.weight+items[v.level].weight;
            v.profit = u.profit+items[v.level].value;
            v.bound = bound(v, nItems, items);
            for(i=0; i<=u.level; i++){
                v.select[i] = u.select[i];
            }
            v.select[v.level] = 1;

            //更新当前最优解
            if(v.profit > maxProfit){
                maxProfit = v.profit;
                for(i=0; i<nItems; i++){
                    solution[i] = v.select[i];
                }
            }

            //如果结点的价值上界大于当前最优解，则加入队列
            if(v.bound > maxProfit){
                enqueue(&q, v);
            }
        }
    }

    return maxProfit;
}

int main(){
    int nItems = 5;
    Item items[5] = {{2, 12, 0}, {1, 10, 0}, {3, 20, 0}, {2, 15, 0}, {5, 25, 0}};
    int capacity = 8;
    int solution[5] = {0};

    int maxProfit = knapsack(nItems, items, capacity, solution);

    printf("Total profit: %d\n", maxProfit);
    printf("Solution: ");
    for(int i=0; i<nItems; i++){
        printf("%d ", solution[i]);
    }
    printf("\n");

    return 0;
}

其中，Item结构体存储物品的重量、价值和价值上界；Node结构体存储结点的决策树层数、当前已获得的价值、当前已占用的重量、价值上界和选择情况；Queue结构体为先进先出队列。在主函数中，定义了5个物品，背包容量为8，使用solution数组存储选中的物品，最终输出了最大价值和选择情况。