0-1背包优化动态规划算法之跳跃点法

// 动态规划 背包问题 跳跃点优化#include <iostream>using namespace std;template<class Type>void Traceback(int n,Type w[],Type v[],Type **p,int *head,int x[]){    Type j = p[head[0]-1][0],m=p[head[0]-1][1];    for(int i=1; i<=n; i++)    {        x[i]=0;        for(int k=head[i+1]; k<=head[i]-1; k++)        {            if(p[k][0]+w[i]==j && p[k][1]+v[i]==m)            {                x[i]=1;                j=p[k][0];                m=p[k][1];                break;            }        }    }}template<class Type>int Knapsack(int n,Type c,Type v[],Type w[],int **p,int x[]){    int *head = new int[n+2];    head[n+1]=0;    p[0][0]=0;    p[0][1]=0;    int left = 0,right = 0,next = 1;    head[n]=1;    for(int i=n; i>=1; i--)    {        int k = left;        for(int j=left; j<=right; j++)        {            if(p[j][0]+w[i]>c) break;            Type y = p[j][0] + w[i],m = p[j][1] + v[i];            while(k<=right && p[k][0]<y)            {                p[next][0]=p[k][0];                p[next++][1]=p[k++][1];            }            if(k<=right && p[k][0]==y)            {                if(m<p[k][1])                {                    m=p[k][1];                }                k++;            }            if(m>p[next-1][1])            {                p[next][0]=y;                p[next++][1]=m;            }            while(k<=right && p[k][1]<=p[next-1][1])            {                k++;            }        }        while(k<=right)        {            p[next][0]=p[k][0];            p[next++][1]=p[k++][1];        }        left = right + 1;        right = next - 1;        head[i-1] = next;    }    Traceback(n,w,v,p,head,x);    return p[next-1][1];}int main(){    int c;    cout<<"请输入待装物品的数量: "<<endl;    int N;    cin>>N;    int w[N+1];    int v[N+1];    cout<<"请输入各个物品的重量及其对应价值:"<<endl;    for(int i=1; i<=N; i++)    {        cin>>w[i];        cin>>v[i];    }    int x[N+1];    cout<<"请输入背包最大载重量:"<<endl;    cin>>c;    int **p = new int *[50];    for(int i=0; i<50; i++)    {        p[i] = new int[2];    }    cout<<"背包能装的最大价值为："<<Knapsack(N,c,v,w,p,x)<<endl;    cout<<"背包装下的物品编号为："<<endl;    for(int i=1; i<=N; i++)    {        if(x[i]==1)        {            cout<<i<<" ";        }    }    cout<<endl;    for(int i=0; i<50; i++)    {        delete p[i];    }    delete[] p;    return 0;}