蛮力法、动态规划法 求解01背包问题

v[i][j]=values[i]+v[i-1][j-weigths[i]];
else
v[i][j]=v[i-1][j];
}
else v[i][j]=v[i-1][j];
}
return v[n][m];
}
int main()
{
int m,n;int i,j;
cout<<"请输入背包的承重量:"<<endl;
2)复杂度分析:2n
2、动态规划法
1)基本思想:Dynamic programming is a technique for solving problems with overlapping subproblems.The function:
V(i,0)=V(0,j)=0;(1)
V(i-1,j)j<w
if (cur_weight <= capacity && cur_value > max_value) {
max_value = cur_value;
}
return;
}
c[d] = 0;
MFKnapsack(capacity, values, weights, c,
d + 1, max_value);
cout << MFKnapsack(capacity, values, weights, n) << endl;
return 0;
}
(2)Dynamic Programming
#include<iostream.h>
#include<string.h>
int v[10][100];//对应每种情况的最大价值
//Output: The value of an optimal feasible subset of the first i items
//Note: Uses as global variables input arrays Weights[1..n],Values[1..n],and table //V[0..n,0..w] whose entries are initialized with-1’s except for row 0 and column 0 //initialized with0’s
cin>>m;
cout<<"请输入物品的总个数:"<<endl;
cin>>n;
cout<<"旅行者背包能装的最大总价值为:"<<MFknapsack(m,n)<<endl;
return 0;
}
五、运行输出结果：
(1)Brute Force：
(2)Dynamic Programming：
1、蛮力法
1)基本思想:Exhaustiving all the combinations of the items, then calculating the maximum values of the items what is not exceed theknapsackcapacity value.
MFKnapsack(capacity,values,weights,c,0,max_value);
return max_value;
}
int main()
{
int capacity=10,n=7;
int values[]={6,3,5,4,6,3,10};
int weights[]={1,2,5,5,4,4,6};
cin>>values[i];
for(i=0;i<10;i++)
for(j=0;j<100;j++)
v[i][j]=0;//初始化数组
for(i=1;i<=n;i++)//递归
for(j=1;j<=m;j++)
{
if(weigths[i]<=j) //如果当前物品的容量小于背包容量
{
if(values[i]+v[i-1][j-weigths[i]]>v[i-1][j])
{
if (d== c.size()) {
int cur_weight = 0, cur_value = 0;
for (int i = 0; i != d; ++i) {
if (c[i] == 1) {
cur_weight += weights[i];
cur_value += values[i];
}
}
(1)Brute Force：
#include <iostream>
#include <vector>
using namespace std;
void MFKnapsack(int capacity, int *values, int *weights,
vector<int> &c, int d, int &max_value)
if V[i,j]<0
if j< Weights[i]
valueMFKnapsack(i-1,j)
else
valuemax(MFKnapsack(i-1,j),Values[i]+MFKnapsack(i-1,j-Weights[i]))
V[i,j]value
return V[i,j]
四、源程序及注释：
c[d] = 1;
MFKnapsack(capacity, values, weights, c,
d + 1, max_value);
}
int MFKnapsack(int capacity, int *values, int *weights, int n)
{
int max_value=0;
vector<int> c(n,0);
V(i,j)=(2)
max{V(i-1,j),V(i-1,j-wi)+vi}j>=w
2)复杂度分析：2n
三、程序伪码：
(1)Brute Force:
Algorithm MFKnapsack(capacity, values, weights,n)
//Input: A nonnegative integer nindicatingthe number of the frist items being //considered and a nonnegative integercapacityindicating the knapsack’s capacity,A arrayvalues[]indicatingthe values of items,A arrayweights[]indicatingtheweightsof items.
//Output: The value of an optimal feasible subset of the first i items
ifd== c.size()
for (int i = 0; i != d; ++i)
if c[i] == 1
cur_weight += weights[i];
cur_value += values[i];
初始化数组ifweigthsij如果当前物品的容量小于背包容量ifvaluesivi1jweigthsivi1jvijvaluesivi1jweigthsi
第八章
班级：*******学号：**********姓名：*****
一、实验内容：
分别用蛮力法、动态规划法 求解0/1背包问题。
二、所用算法的基本思想及复杂度分析：
int MFknapsack(int m,int n)
{
int i,j,weigths[10],values[10];
cout<<"请输入每个物品的重量："<<endl;
for(i=1;i<=n;i++)
cin>>weigths[i];
cout<<"请输入每个物品的价值："<<endl;
for(i=1;i<=n;i++)
if cur_weight <= capacity && cur_value > max_value
max_valuecur_value;
rming:
Algorithm MFKnapsack(i,j)
//Implements the memory function method for the knapsack problem
//Input: A nonnegative integeri indicatingthe number of the frist items being //considered and a nonnegative integer j indicating the knapsack’s capacity