0-1背包问题简单实例（动态规划求…

    适用动态规划对0-1背包问题进行了简单的描述，容易看，也可以直接用

bag1.cpp：

#include <iostream>

using namespace std;

void Knapsack(int *v, int *w, int c, int n, int **m);

int main()

{

//0-1 bag problem, using dynamic programming

int c;//weight constrain of the bag

  int *w;//weight of goods

int *v;//value of goods

int n;//kinds of goods

int **m;//m(i,j) means optimal value of i~n goods under weight constrain j

int i, j;

cout<<"Please input weight constrain of the bag:";

cin>>c;

cout<<"Please input number of kinds of goods:";

cin>>n;

if(n <= 0 || c <= 0)

{

cout<<"invalid arguments"<<endl;

return 0;

}

cout<<"Please input "<<n<<" weight of goods:"<<endl;

w = new int[n + 1];

w[0] = 0;

for(i = 1; i <= n; i++)

{

cin>>w[i];

}

cout<<"Please input "<<n<<" value of goods:"<<endl;

v = new int[n + 1];

v[0] = 0;

for(i = 1; i <= n; i++)

{

cin>>v[i];

}

m = new int *[n + 1];

for(i = 0; i <= n; i++)

{

m[i] = new int[n + 1];

}

for(i = 0; i <= n; i++)

{

for(j = 0; j <= n; j++)

{

m[i][j] = 0;

}

}

Knapsack(v, w, c, n, m);

cout<<"The optimal value is: "<<m[1][c]<<endl;

delete m;

delete w;

delete v;

return 0;

}

void Knapsack(int *v, int *w, int c, int n, int **m)

{

int i, j;

int jMax = min(w[n] - 1, c);

for(j = 0; j <= jMax; j++)

{

m[n][j] = 0;

}

for(j = w[n]; j <= c; j++)

{

m[n][j] = v[n];

}

for(i = n - 1; i > 1; i--)

{

jMax = min(w[i] - 1, c);

for(j = 0; j < jMax; j++)

{

m[i][j] = m[i + 1][j];

}

for(j = w[i]; j <= c; j++)

{

m[i][j] = max(m[i + 1][j - w[i]] + v[i], m[i + 1][j]);

}

}

m[1][c] = m[2][c];

if(c >= w[1])

{

m[1][c] = max(m[1][c], m[2][c - w[1]] + v[1]);

}

}

Makefile:

bag1:bag1.o

g++ -o bag1 bag1.o

bag1.o:bag1.cpp

g++ -c bag1.cpp

运行结果：

Please input weight constrain of the bag:10

Please input number of kinds of goods:5

Please input 5 weight of goods:

2

2

6

5

4

Please input 5 value of goods:

6

3

5

4

6

The optimal value is: 15