分支限界算法求解0-1背包問題：不構造最優解，使用struct和queue

#include<iostream>
#include<queue>
using namespace std;
#define NUM 100

struct Node
{
	int weight;		//重量 
	int value;		//價值 
	int level;		//層次 
	int flag;		//1是左孩子，0是右孩子 
};
queue<Node> q;

//增加結點 
int enQueue(int w,int v,int level,int flag,int n,int* best)
{
	Node node;
	node.weight = w;
	node.value = v;
	node.level = level;
	node.flag = flag;
	if (level == n)
	{
		if (node.value > *best)
			*best = node.value;
		return 0;
	}
	else  q.push(node);
}
//w為重量數組，v為價值數組，n為物品個數，c為背包容量，best為全局最大價值
int knapsack(int w[],int v[],int n,int c,int* best)
{
	int i=1;			//第一個結點 
	Node live;
	live.weight = 0;
	live.value = 0;
	live.level = 1;
	live.flag = 0;
	Node tmp;			//根結點 
	tmp.weight = -1;	//層分界标志 
	tmp.value = 0;
	tmp.level = 0;
	tmp.flag = 0;
	q.push(tmp);
	while (true)
	{
		if (live.weight + w[i-1] <= c)
		{
			enQueue(live.weight + w[i-1], live.value + v[i-1], i, 1,n,best);
		}
		enQueue(live.weight,live.value, i, 0,n,best);
		live = q.front();
		q.pop();
		if (live.weight == -1)
		{
			if (q.empty() == 1) break;
			live = q.front();
			q.pop();
			q.push(tmp);
			i++;
		}
	}
	return 0;
}
int main()
{
	int w[NUM];
	int v[NUM];
	int c;
	int n;
	while(scanf("%d%d",&c, &n)!=EOF)
	{
		for(int i=0; i<n; i++)
			scanf("%d%d",&w[i],&v[i]);
		int best=0;
		knapsack(w,v,n,c,&best);
		printf("%d\n", best);
	}
	return 0;
}