蛮力法与动态规划法求解0/1背包问题

蛮力法与动态规划法求解0/1背包问题

蛮力法（我用的是穷举2的n方个可能选择）

#include
#include 
#include 
#include
#define N 100
using namespace std;

void conversion(int n,int b[]){
	int i;
	for(i=0;i<N;i++){
	b[i] = n%2;
	n = n/2;
	if(n==0)
		break;
	}
}

int *BagProblem(int w[],int v[],int C,int len){
	int maxValue = 0;
	int b[N];
	int *temp;//temp就是解
	temp = (int *)malloc(sizeof(int)*len); 
	int value,weight,i,j;
	
	//穷举2的n方个可能的选择 
	for(i=0;i<pow(2,len);i++){
		//初始化数组b为0 
		for(j=0;j<len;j++){
			b[j] = 0;
		}
		//转化 
		conversion(i,b);
		value = 0;
		weight = 0;
		for(j=0;j<len;j++){
			weight += w[j]*b[j];
			value += v[j]*b[j];
		}
		//如果当前的子集满足此两个条件，则为目前最优 
		if(weight<=C&&maxValue<value){
			maxValue = value;
			//将temp重置 
			for(j=0;j<len;j++){
				temp[j]=0;
			}
			//将当前的b[j]复制到temp保存 
			for(j=0;j<len;j++){
				temp[j]=b[j];
			}
		} 
	}
	//循环完毕，返回结果子集
	return temp;
}

int main(){
	int w[]={2,2,6,5,4};
	int v[]={6,3,5,4,6};
	int C = 10,value = 0;
	int len = 5,i;
	
	//求时间花费 
	LARGE_INTEGER nFreq;
    LARGE_INTEGER nBeginTime;
    LARGE_INTEGER nEndTime;
    double time;
	QueryPerformanceFrequency(&nFreq);
	QueryPerformanceCounter(&nBeginTime);
	//计时代码区间
	 
	int *result = BagProblem(w,v,C,len);
	
	QueryPerformanceCounter(&nEndTime);
	time=(double)(nEndTime.QuadPart-nBeginTime.QuadPart)*1000000000/(double)(nFreq.QuadPart);
	//不计时代码区间 
	for(i=0;i<len;i++){
		if(result[i]==0)
			continue;
		else
			cout<<i+1<<" ";//输出所选物品的序号 
	}
	cout<<endl;
	for(i=0;i<len;i++){
		value += v[i]*result[i];
	}
	cout<<"取得的最大价值为:"<<value<<endl;
    cout  << time; //单位是纳秒.
	
	return 0;
}

动态规划法

#include
#include 
#include
using namespace std;

//动态规划法求解0/1背包问题，实质是一个填表的过程 
int *KnapSack(int w[],int v[],int n,int C){
	int V[n+1][C+1];
	int *x;
	x = (int *)malloc(sizeof(int)*(n-1));
	int i,j;
	//初始化第0列 
	for(i=0;i<=n;i++)
		V[i][0] = 0;
	//初始化第0行 
	for(j=0;j<=C;j++)
		V[0][j] = 0;
	//计算第i行，进行第i次迭代
	for(i=1;i<=n;i++)
		for(j=1;j<=C;j++)
			if(j<w[i])
				V[i][j] = V[i-1][j];
			else
				V[i][j] = max(V[i-1][j],V[i-1][j-w[i]]+v[i]);
	//求放入背包的物品 
	for(j=C,i=n;i>0;i--){
		if(V[i][j]>V[i-1][j]){
			x[i]=1;
			j=j-w[i];
		}
		else
			x[i]=0;
	}
	return x;//输出所选物品 
}

int main(){
	int w[]={0,2,2,6,5,4};
	int v[]={0,6,3,5,4,6};
	int C = 10;
	int len = 5,i,value = 0;
	
	//求时间花费 
	LARGE_INTEGER nFreq;
    LARGE_INTEGER nBeginTime;
    LARGE_INTEGER nEndTime;
    double time;
	QueryPerformanceFrequency(&nFreq);
	QueryPerformanceCounter(&nBeginTime);
	//计时代码区间
	 
	int *result = KnapSack(w,v,len,C);
	
	QueryPerformanceCounter(&nEndTime);
	time=(double)(nEndTime.QuadPart-nBeginTime.QuadPart)*1000000000/(double)(nFreq.QuadPart);
	//不计时代码区间 
	for(i=1;i<=len;i++){
		if(result[i]==0)
			continue;
		else
			cout<<i<<" ";//输出所选物品的序号 
	}
	cout<<endl;
	for(i=1;i<=len;i++){
		value += v[i]*result[i];
	}
	cout<<"取得的最大价值为:"<<value<<endl;
    cout  << time; //单位是纳秒.
	
	return 0;
}