背包问题（Knapsack problem）之01

问题描述：

给定一组物品，每种物品都有自己的重量和价格，在限定的总重量内，我们如何选择，才能使得物品的总价格最高。

转移方程：dp[i][j] = max(dp[i-1][j],dp[i-1][j-weight[i]] + value[i]）

不放 放入

其中dp[i][j]表示放入i个物品，总价值为j。

代码如下：

// beibaoproblem.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>

using namespace std;

#define max(a, b) ( (a) > (b) ? (a) :(b) )

int _tmain(int argc, _TCHAR* argv[])
{

const int v = 10;  //最大容量
const int n = 3;  //物件的个数

int value[] = {4,5,6};
int weight[] = {3,4,5};
int temp1, temp2;

int i,j;

int dp[n + 1][v + 1];

//初始化
for (i = 0; i < n+1; i++)
for (j = 0; j < v + 1; j++)
{
dp[i][j] = 0;
}

for(i = 1; i <= n; i++)
{
for (j = 1; j <= v; j++)
{
if (j >= weight[i-1])//表示剩余容量大于第i件的容量，可以放入
{

dp[i][j] = max(dp[i-1][j], dp[i-1][j - weight[i-1]] + value[i-1]);

}
else              //反之不放入
{
dp[i][j] =  dp[i-1][j];

}
}

}

for (i = 0; i < n+1; i++)
{
for (j = 0; j < v + 1; j++)
{
cout << dp[i][j]<<" " ;
}
cout << endl;
}

cout <<  dp
[v];

return 0;
}

结果：



-----------------------------------------------------------------------------------------------------

以下是从文件读取:

// beibaoproblem.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>

using namespace std;

#define FILENAMELENGTH 1000
#define max(a,b) (a) > (b) ? a : b

class CBeibao
{
public:
int m_nNumber;     //物品数量
int m_nMaxWeight;  //最大载重量

int *m_pWeight;  //每个物品的重量
int *m_pValue;   //每个物品的价值

int *m_pCount;   //每个物品被选中的次数
int m_nMaxValue; //最大价值

public:
CBeibao(const char* filename);
~CBeibao();

int GetMaxValue();
//n表示物品个数，m背包载重量，w数组重量，v价值数组，c是否被选中数组
int GetMaxValue(int n, int m, int *w, int *v, int *c);
void Display(int nmaxValue);
void Display(int nMaxValue, const char* filename);

};

//读入数据
CBeibao::CBeibao(const char* filename)
{
FILE *fp;
fopen_s(&fp, filename, "r");
if (fp == NULL)
{
printf_s("can not open file");
return;
}

fscanf_s(fp, "%d%d", &m_nNumber, &m_nMaxWeight);

m_pWeight = new int[m_nNumber+1];
m_pValue = new int[m_nNumber+1];

//读入每个物品的重量
m_pWeight[0] = 0;
for(int i = 1; i <= m_nNumber; i++)
fscanf_s(fp, "%d", m_pWeight+i);

//读入每个物品的价值
m_pValue[0] = 0;
for(int i = 1; i <=m_nNumber;i++)
fscanf_s(fp, "%d", m_pValue+i);

//初始化每个物品被选中次数为0
m_pCount = new int[m_nNumber+1];
for(int i = 0; i <=m_nNumber;i++)
m_pCount[i] = 0;

fclose(fp);

}

CBeibao::~CBeibao()
{
delete[] m_pWeight;
m_pWeight = NULL;

delete[] m_pValue;
m_pValue = NULL;

delete[] m_pCount;
m_pCount = NULL;

}

int CBeibao::GetMaxValue(int n, int m, int *w, int *v, int *c)//n表示物品个数，m背包载重量，w数组重量，v价值数组，c是否被选中数组
{
int row = n+1;
int col = m+1;

int i,j;

//value[i][j]表示前i个物品能装入载重量为j的背包中物品的最大价值
int **value = new int *[row];
for(i = 0; i < row; i++)
value[i] = new int[col];

//初始化为0
for(i = 0; i < row; i++)
for(j = 0; j < col; j++)
value[i][j] = 0;

//计算
for(i = 1; i < row; i++)
for (j = 1; j < col; j++)
{
if (j >= w[i])
{
value[i][j] = max(value[i-1][j], value[i-1][j - w[i]] + v[i]);
}
else
{
value[i][j] = value[i-1][j];
}
}

//逆推求装入的物品
j = m;
for (i = row - 1; i > 0; i--)
{
if (value[i][j] > value[i-1][j])  //表示第i个物品装入
{
c[i] = 1;
j -= w[i];
}
}

//记录最大价值
int nMaxValue = value[row-1][col-1];

//释放该二维数组
for (i = 0; i < row; i++)
{
delete[col] value[i];
value[i] = NULL;
}

delete[] value;
value = NULL;

return nMaxValue;

}

int CBeibao::GetMaxValue()
{
int nValue = GetMaxValue(m_nNumber, m_nMaxWeight, m_pWeight, m_pValue, m_pCount);
m_nMaxValue = nValue;
return nValue;
}

//显示结果
void CBeibao::Display(int nMaxValue)
{
_tprintf(_T(" %d"), nMaxValue);
for(int i = 1; i <= m_nNumber; i++)
{
if (m_pCount[i])
_tprintf(_T(" %d %d"), i, m_pCount[i]);
}
_tprintf(_T(""));
}

void CBeibao::Display(int nMaxValue, const char* filename)
{
FILE *fp;
fopen_s(&fp, filename, "w");
if (fp == NULL)
{
_tprintf(_T("can not write file!"));
return;
}

fprintf(fp, "%d", nMaxValue);
for (int i = 1; i <= m_nNumber; i++)
{
if (m_pCount[i])
{
fprintf(fp, "%d %d", i, m_pCount[i]);
}
}

fclose(fp);
}

int _tmain(int argc, _TCHAR* argv[])
{
char sinput[10];
char sfilename[FILENAMELENGTH] = "C:\\Users\\sony\\Desktop\\k\\practice\\beibaoproblem\\beibaoproblem\\input.txt";

scanf_s("%s", sinput, _countof(sinput));

while (_stricmp(sinput, "q") != 0)
{
if (_stricmp(sinput, "i") == 0)
{
_tprintf(_T(" please input a filename:"));
/*scanf_s("%s", sfilename);*/

//获取满足最大载重量的最大价值
CBeibao beibao(sfilename);

int nMaxValue = beibao.GetMaxValue();
if (nMaxValue)
{
beibao.Display(nMaxValue);
int nlen = strlen(sfilename);
//这里sfilename表示头指针，sttrcpy表示复制到sfilename + nlen - 4
strcpy_s(sfilename + nlen - 4, FILENAMELENGTH , "_result.txt");
beibao.Display(nMaxValue, sfilename);

}
else
{
_tprintf(_T("error!"));
}
}

_tprintf(_T("input command: "));
scanf_s("%s", sinput);

}

return 0;
}

参考链接：http://blog.csdn.net/livelylittlefish/article/details/2186206
http://www.cnblogs.com/usa007lhy/archive/2013/05/19/3087195.html