简单的几道算法题_AboutDP

#include<iostream>
using namespace std;


/**
整数切分。比如2有1种1+1
3有2种 1+1+1  2+1
4有4种 1+1+1+1 2+2  1+3 1+1+2
5有6种 1+1+1+1+1 1+1+3 1+1+1+2 2+2+1 2+3 1+4
*/
int IntergeSplit(int n)
{
	if (n <= 1)
		return 0;
	int **dp = new int*[n+1];
	for(int i = 0; i <= n;++i)
		dp[i] = new int[n+1];

	dp[0][0] = 1;
	for(i = 1; i <= n;++i)
		dp[i][0] = 0;

	for(int j = 1; j <= n;++j)
		dp[0][j] = 0;

	for(i = 1; i <= n; ++i)
	{
		for(j = 1; j <= n;++j)
		{
			dp[i][j] = 0;
			for(int k = 0; k <= j; ++k)
			{ 
				if( k > i-j )
					break;
				dp[i][j] += dp[i-j][k];
			}
			/*
			int number_j = i/j; //在整数i的切分里，最多有number_j个 j
			for(int k = 1; k <= number_j; ++k)
			//	dp[i][j] += dp[i-k*j][j];
				for(int m = 0; m < j; ++m)
					dp[i][j] += dp[i-k*j][m];
					*/
		}
	}

	int result = 0;
	for(j = 1; j <= n; ++j)
		result += dp[n][j];

	for(i = 0; i<= n;++i)
		delete []dp[i];
	delete []dp;
//	delete []*dp;
//	delete []dp;
	return result - 1;

}

//0-1背包问题
int max(int a,int b)
{
	return a > b?a:b;
}
void Packet0_1(int W,int w[],int value[],unsigned int length)
{
	if(W == 0)
		return;
	int **dp = new int*[length+1];
	for(int i = 0; i <= length; ++i)
		dp[i] = new int[W+1];

	for(i = 0; i <= length; ++i)
		dp[i][0] = 0;
	for(i = 0; i <= W; ++i)
		dp[0][i] = 0;

	//j 是当前背包容量
	//i 是当前可选的前i个物品
	for(i = 1; i <= length; ++i)
	{
		for(int j = 1; j <= W; ++j)
		{
			//当前容量是否能放下物品j 
			//w[]下标从0开始,因此都要i-1,value[]也是如此
			if( j >= w[i-1])
			{
				//看加入物品i与不加入物品i，哪个价值大。
				dp[i][j] = max(dp[i-1][j-w[i-1]]+value[i-1],dp[i-1][j]);
			}
			else
				dp[i][j] = dp[i-1][j];
		}
	}
	cout<<dp[length][W]<<endl;

	for(i = 0; i <=length; ++i)
		delete [] dp[i];
	delete []dp;
}
void main()
{
	/*
	for(int i = 0; i <= 20; i++)
		cout<<i<<"    "<<IntergeSplit(i)<<endl;
*/

	int w[] = {3,4,5,6,10};
	int v[] = {1,2,4,3,5};
	int len = sizeof(w)/sizeof(int);
	Packet0_1(10,w,v,len);


}