回顾基本01背包问题

今天练了一下当年的背包问题，发现居然OJ还没过，赶紧刷一下。想到当年，每次做背包，就和军爷一起回忆
Eason的你的背包，从此变得好浪漫~~~~~


题目太经典了，分析就略过了。写个递推方程：
c[i,j]=max{c[i,j-1], c[i-1,j-w[i]+v[i]]} if BagWeight>=j>=w[i]
      =c[i,j-1]  if  j<w[i]   
c[1,j]={0  0<=j<w[i]   v[1]   w[1]<j<=BagWeight}
c[i,0]=0  i=1,2...n
 
coding过程中遇到一个问题，就是背包数n肯定是离散的，1,2,3.。n，然而背包重量BagWeight是离散的么，循环的step怎么设置，
后来仔细看了下，背包问题的包重和价值都是整数，所以其实是一个整数规划，是一类问题，不考虑小数，所以才可以BagWeight那一维
也是step为1递推。


还有一个重要的问题，好多次都忽视了，就是
初始化c[1,j]={0 if 0<=j<w[i]    v[1]   if  w[1]<=j<=bagweight}      (w[1] < = BagWeight)
这么写的前提是 w[1] < = BagWeight, 所以需要考虑另一种情况
c[1,j]={0 if 0<=j<=BagWeight}   (w[1] > BagWeight)


递推出口尚且如此，自然递推通项式也是如此，所以一开始看到教材里第二层loop 是需要用个min函数其实是殊途同归的。
所以大家根据这里的分析是需要改递推方程的，要扩充成两种情况，小生此处就不扩了，一切尽在代码里，而且书上也没扩~~~

另外还有就是此题还要知道解的路径，思路是根据c[n][BagWeight] 递推回去，看等于哪个，就赋值为选或者不选该项。
此题有个刮三地方就是需要返回所有解中对应的二进制数最小的，我大概当时恰在这里了，现在想想其实就是
if c[i][j]=c[i-1][j] || c[i-1][j-w[i]+v[i]], select latter, as it will lead right more 1, and left more 0, which lead small binary number
等于的话优先走1的那条回溯路径，就对了这样前面就0了，是解中二进制数最小的。


附上代码：

#include <iostream>
using namespace std;
int ZeroOneBag(int**c, bool* Select, int* Value, int* Weight, int n, int BagWeight)
{
	//array start with 1, length is n+1
	for(int i=1;i<=n;i++)
		c[i][0]=0;
	if(Weight[1]<=BagWeight)
	{
		for(int j=0;j<Weight[1];j++)//j<Weight[1] means can not add bag 1 into bag
			c[1][j]=0;
		for(int j=Weight[1];j<=BagWeight;j++)
			c[1][j]=Value[1];
	}
	else
	{
		for(int j=0;j<=BagWeight;j++)//j<Weight[1] means can not add bag 1 into bag
			c[1][j]=0;
	}




	for(int i=2;i<=n;i++)
	{
		if(Weight[i]<=BagWeight)
		{
			for(int j=0;j<Weight[i];j++)
				c[i][j]=c[i-1][j];
			for(int j=Weight[i];j<=BagWeight;j++)
				c[i][j]=max(c[i-1][j],c[i-1][j-Weight[i]]+Value[i]);
		}
		else
		{
			for(int j=0;j<=BagWeight;j++)
				c[i][j]=c[i-1][j];
		}
	}
	int i=n, BagWeightTmp=BagWeight;
	while(i>=2)
	{
		if(BagWeightTmp>=Weight[i])
		{
			if(c[i][BagWeightTmp]==(c[i-1][BagWeightTmp-Weight[i]]+Value[i]))//if c[i][j]=c[i-1][j] || c[i-1][j-w[i]+v[i]], select latter, as it will lead right more 1, and left more 0, which lead small binary number
			{
				Select[i]=true;
				BagWeightTmp-=Weight[i];
			}
			else if(c[i][BagWeightTmp]==c[i-1][BagWeightTmp])
				Select[i]=false;
			else
				;
		}
		else
		{
			Select[i]=false;
		}
		i--;
		
	}
	if(BagWeightTmp>=Weight[1])
		Select[1]=true;
	else
		Select[1]=false;
	return c[n][BagWeight];
}


int main()
{
	int BagWeight, n;
	int casei=1;
	while(cin>>n>>BagWeight)
	{
		int BestBagValue;
		int *Weight=new int[n+1]();
		int *Value=new int[n+1]();
		bool * Select=new bool[n+1]();


		int **c=new int*[n+1];
		for(int i=0;i<=n;i++)
			c[i]=new int[BagWeight+1]();




		for(int i=1;i<=n;i++)
			cin>>Value[i];
		for(int i=1;i<=n;i++)
			cin>>Weight[i];
		BestBagValue=ZeroOneBag(c,Select,Value,Weight,n,BagWeight);
		cout<<"Case "<<casei<<endl;
		casei++;
		cout<<BestBagValue<<" ";
		for(int i=1;i<=n;i++)
			if(Select[i]==false)
				cout<<"0";
			else
				cout<<"1";
		cout<<endl;


		for(int i=0;i<=n;i++)
			delete[] c[i];
		delete[] c;




		delete[] Select;
		delete[] Value;
		delete[] Weight;
	}
	return 0;
}

恩 还碰到了一个申请c[][]的问题，居然申请了n*n得数组，于是申请的堆出了一个奇怪的bug，果然从coding开始每一句都要知道自己再干啥，记得之前好像在哪里看到过这句话。。。。而且一定要集中，我好像在这步开了小菜