算法学习-（二）动态规划算法

动态规划算法

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例1、

#include<iostream>
using namespace std;

/*
*例1：求用1*2的瓷砖覆盖2*M的地板有几种方式？
*/

int Func(int M);//分治法
int Func1(int M);//动态规划法（备忘录法）
int Func2(int M);//动态规划法,自底向上
int main()
{
	int n;
	while(cin >> n)
		cout << Func2(n) << endl;
	return 0;
}
int Func2(int M)
{
	int a[1000] = {0};
	int i=0;
	a[1]=1;
	a[2]=2;
	for(i=3; i<=M; i++)
		a[i] = a[i-1] + a[i-2];
	return a[M];
}
int Func1(int M)
{
	int a[1000]={0};
	//assert(M>0);
	if(a[M] != 0)
		return a[M];
	if(M == 1)
		a[M] = 1;
	else if(M == 2)
		a[M] = 2;
	else
	{
		int b0 = Func1(M-1);
		int b1 = Func1(M-2);
		a[M] = b0 + b1;
	}	
	return a[M];
}
int Func(int M)
{
	if(M == 0)
		return 0;
	else if(M == 1)
		return 1;
	else if(M == 2)
		return 2;
	else
		return Func(M-1)+Func(M-2);
}

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例2、LCS问题

/*LCS最长公共子序列问题*/
#define N 100
#define INF 9999999
int LCS_LENGTH(const char* X, const char* Y);//动态规划算法，自底向上
int LCS_LENGTH1(const char* X, const char* Y);//动态规划算法的变形（备忘录法），自顶向下
int LCS_Memo(const char* X, const char* Y, int m, int n, int a[][N]);
//int a[N][N];

int main()
{
	char *X = "ABCBDAB";
	char *Y = "BDCABA";
	cout << LCS_LENGTH1(X, Y) << endl;
	return 0;
}
int LCS_LENGTH1(const char* X, const char* Y)
{
	int a[N][N]={0};
	int m = strlen(X);
	int n = strlen(Y);
	//memset(a, INF, sizeof(a));
	LCS_Memo(X, Y, m, n, a);
	return a[m][n];
}

int LCS_Memo(const char* X, const char* Y, int m, int n, int a[][N])
{

	//if(a[m][n] < INF)
	if(a[m][n] != 0)
		return a[m][n];
	if(m==0 || n==0)
		a[m][n]=0;
	else if(*(X + m) == *(Y + n))
		a[m][n] = LCS_Memo(X, Y, m-1, n-1, a) + 1;
	else
	{
		int p = LCS_Memo(X, Y, m-1, n, a);
		int q = LCS_Memo(X, Y, m, n-1, a);
		int max = p > q ? p : q;
		a[m][n] = max;
	}
	return a[m][n];
}

int LCS_LENGTH(const char* X, const char* Y)
{
	int a[N][N] = {0};
	int m = strlen(X);
	int n = strlen(Y);
	int i, j;
	if(X == NULL || Y == NULL)
		return 0;
	for(i=1; i<=m; i++)
	{
		for(j=1; j<=n; j++)
		{
			if(*(X+i-1) == *(Y+j-1))
				a[i][j] = a[i-1][j-1]+1;
			else
			{
				int max = a[i-1][j] > a[i][j-1] ? a[i-1][j] : a[i][j-1];
				a[i][j] = max;
			}	
		}
	}
	return a[m][n];
}

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例3、0-1背包问题

代码：

#include<iostream>
#include<string.h>
using namespace std;
#define N  3
#define V  5

int bagZO(int W);
int main()
{
	int C[4] = {0, 1, 2, 3};
	int W[4] = {0, 60, 100, 120};
	int f[N+1][V+1] = {0};
	int i, j;
	for(i=1; i<=N; ++i)
	{
		for(j=1; j<=V; ++j)
		{
			if(j<C[i])
				f[i][j]=f[i-1][j];
			else
				f[i][j]=f[i-1][j] > (f[i-1][j-C[i]]+W[i]) ? f[i-1][j] : (f[i-1][j-C[i]]+W[i]);
		}
	}
	cout << f[N][V] << endl;
	return 0;
}