C++回溯法求0-1背包问题

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我是靠谱客的博主痴情白昼，这篇文章主要介绍C++回溯法求0-1背包问题，现在分享给大家，希望可以做个参考。

主要思想：先将数组w和数组p按照单价进行排序，利用结构体的index保存其下标。

bound函数是求当前最大可能价值。

backtrack函数是利用回溯法，如果增加当前物品，则想x[i]=1，否则为0。当i>n时，递归调用结束，并且更新数组bestx和bestp。

复制代码

#include <iostream>
#include <algorithm>
#define N 4
using namespace std;
int c = 7, w[N+1] = {0,3,5,2,1}, p[N+1]={0,9,10,7,4}, cw = 0, cp = 0, bestp = 0, bestx[N+1] = {0}, x[N+1] = {0};

struct node
{
    double v;
    int index;
}no[N+1];

bool cmp(struct node n1, struct node n2)
{
    return n1.v > n2.v;
}

void sort_value()
{
    for(int i = 1; i <= N; i++)
    {
        no[i].v = 1.0*p[i]/w[i];
        no[i].index = i;
    }
    sort(no+1,no+N,cmp);
}

double bound(int i)
{
    double cleft = c - cw;
    double bound = cp;
    while(i <= N && w[no[i].index] <= cleft)
    {
        cleft -= w[no[i].index];
        bound += p[no[i].index];
        i++;
    }
    if(i <= N)
        bound += p[no[i].index] / w[no[i].index] * cleft;
    return bound;
}

void backtrack(int i)
{
    if(i > N)
    {
        if(cp > bestp)
        {
            for(int j = 1; j <= N; j++)
                bestx[no[j].index] = x[no[j].index];
            bestp = cp;
        }
        return;
    }
    if(cw + w[no[i].index] <= c)
    {
        x[no[i].index] = 1, cw += w[no[i].index], cp += p[no[i].index];
        backtrack(i+1);
        cw -= w[no[i].index], cp -= p[no[i].index];
    }
    if(bound(i+1) > bestp)
    {
        x[no[i].index] = 0;
        backtrack(i+1);
    }
}

int main()
{
    sort_value();
    backtrack(1);
    for(int i = 1; i <= N; i++)
        cout << bestx[i] << ' ';
    cout << endl;
    cout << bestp << endl;
    return 0;
}

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#include <iostream>
#include <algorithm>
#define N 4
using namespace std;
int c = 7, w[N+1] = {0,3,5,2,1}, p[N+1]={0,9,10,7,4}, cw = 0, cp = 0, bestp = 0, bestx[N+1] = {0}, x[N+1] = {0};

struct node
{
    double v;
    int index;
}no[N+1];

bool cmp(struct node n1, struct node n2)
{
    return n1.v > n2.v;
}

void sort_value()
{
    for(int i = 1; i <= N; i++)
    {
        no[i].v = 1.0*p[i]/w[i];
        no[i].index = i;
    }
    sort(no+1,no+N,cmp);
}

double bound(int i)
{
    double cleft = c - cw;
    double bound = cp;
    while(i <= N && w[no[i].index] <= cleft)
    {
        cleft -= w[no[i].index];
        bound += p[no[i].index];
        i++;
    }
    if(i <= N)
        bound += p[no[i].index] / w[no[i].index] * cleft;
    return bound;
}

void backtrack(int i)
{
    if(i > N)
    {
        if(cp > bestp)
        {
            for(int j = 1; j <= N; j++)
                bestx[no[j].index] = x[no[j].index];
            bestp = cp;
        }
        return;
    }
    if(cw + w[no[i].index] <= c)
    {
        x[no[i].index] = 1, cw += w[no[i].index], cp += p[no[i].index];
        backtrack(i+1);
        cw -= w[no[i].index], cp -= p[no[i].index];
    }
    if(bound(i+1) > bestp)
    {
        x[no[i].index] = 0;
        backtrack(i+1);
    }
}

int main()
{
    sort_value();
    backtrack(1);
    for(int i = 1; i <= N; i++)
        cout << bestx[i] << ' ';
    cout << endl;
    cout << bestp << endl;
    return 0;
}