学堂在线-程序设计基础-第六章
作者：互联网
文章目录
第六章 递推与动态规划
6.1 兔子数列问题
先设定递推初值；再按照公式进行递推（常用循环实现）。
代码实现
//斐波那契数列
#include <iostream>
using namespace std;
int main()
{
int n;
cout << "求第几个月的兔子个数： " ;
cin >> n ;
int dang_yue = 1;
int shang_yue = 1;
int qian_yue = 1;
for (int i = 2; i < n; i ++)
{
dang_yue = qian_yue + shang_yue;
qian_yue = shang_yue;
shang_yue = dang_yue;
}
cout << "第" << n << "个月的兔子总数为：" << dang_yue << endl;
return 0;
}
6.2 分鱼问题
6.2.2 从 A 到 E 递推-代码实现
//分鱼问题
#include <iostream>
using namespace std;
int main()
{
int num[5]={0};
for (num[0] = 1; ; num[0]++)
{
if (num[0] % 5 != 1)
continue;
int i;
for (i = 1; i <= 4; i++)
{
num[i] = num[i-1] / 5 * 4;
if (num[i] % 5 != 1)
break;
}
if (i <= 4)
continue;
break;
}
for(int i = 0; i < 5; i++)
cout << num[i] << " " << endl;
return 0;
}
6.2.3 从 E 到 A 递推-代码实现
通过逻辑分析，优化枚举初值和枚举步长，提高代码运行效率
//分鱼问题
#include <iostream>
using namespace std;
int main()
{
int num[5]={0};
for (num[0] = 16; ; num[0]+=20)
{
int i;
for (i = 0; i <4; i++)
{
if (num[i]%4!=0)
break;
num[i+1]=num[i]/4*5+1;
}
if (i <4)
continue;
break;
}
for(int i = 0; i < 5; i++)
cout << num[i] << " " << endl;
return 0;
}
不用数组辅助-代码实现
不用数组，反而没法用循环简化代码，不知道有没有其他更好的实现方式。
//分鱼问题
#include <iostream>
using namespace std;
int main()
{
int num_a,num_b,num_c,num_d,num_e=0;
for (num_e = 16; ; num_e+=20)
{
if (num_e%4!=0)
continue;
num_d=num_e/4*5+1;
if (num_d%4!=0)
continue;
num_c=num_d/4*5+1;
if (num_c%4!=0)
continue;
num_b=num_c/4*5+1;
if (num_b%4!=0)
continue;
num_a=num_b/4*5+1;
break;
}
cout << num_a << "\n" << num_b << "\n"
<< num_c << "\n" << num_d << "\n"
<< num_e << endl;
return 0;
}
6.3 橱窗的插画问题
求最优值问题：
- 首先用枚举法，枚举出所有的情况，对于每个情况计算美感和并保存最大值；
- 由于枚举法，每次计算中有许多重复计算，所以用一个数组存储一些部分插花方案的美感值，采用递归法计算，可以省去一部分计算，但是由于定义了一个数组，需要额外存储空间；
- 采用动态规划，计算并存储部分最优方案，极大优化了存储空间的占用；
6.3.3 用枚举思想解题
//橱窗插花，用枚举思想解题
#include <iostream>
using namespace std;
const int V = 5, F = 3;
//把一个数num表示为二进制数组binary[]
void ToBinary(int num, int binary[V])
{
for(int i = 0; i < V; i++)
{
binary[i] = num & 1; //按位与，取最低为
num = num >> 1; //右移一位
}
}
//计算binary中1的个数
int count1(int binary[V])
{
int count = 0;
for (int i = 0; i < V; i++)
count += binary[i];
return count;
}
int main()
{
int beauty[V][F] = {{7 , 5 , -21},
{23 , 21 , 5 },
{-5 , -4 , -4 },
{-24, 10 , -20},
{16 , 23 , 20 }};//定义美感数组
int best_beauty = 0, best_put = 0;//定义最大美感得分，和插花方案
//枚举，从0到（2^V-1）
for (int i = 0; i < (1<<V); i++)
{
int binary[V] = {0};
ToBinary(i, binary);//得到插花方案binary[]
if (count1(binary) != F)//花的数量不对，枚举下一个方案
continue;
//计算美感得分
int sum_beauty = 0;
for (int vase = 0, flower = 0; vase < V; vase++)
if (binary[vase] == 1)//该花瓶中插的有花
{
sum_beauty += beauty[vase][flower];
flower++;
}
//保存最大美感和插花方案
if (sum_beauty > best_beauty)
{
best_beauty = sum_beauty;
best_put = i;
}
}
//输出答案
cout << "最大美感得分：" << best_beauty << endl;
cout << "插花方案：";
int best_binary[V] = {0};
ToBinary(best_put, best_binary);
for (int vase = 0, flower = 1; vase < V; vase++)
{
if (best_binary[vase] == 1)
{
cout << flower;
flower++;
}
else
cout << 0;
}
return 0;
}
6.3.4 采用递归的优化算法
//橱窗插花，采用递归的优化算法
#include <iostream>
using namespace std;
const int V = 5, F = 3;
//把一个数num表示为二进制数组binary[]
void ToBinary(int num, int binary[V])
{
for(int i = 0; i < V; i++)
{
binary[i] = num & 1; //按位与，取最低为
num = num >> 1; //右移一位
}
}
//计算binary中1的个数
int count1(int binary[V], int &high)
{
int count = 0;
high = -1;
for (int i = 0; i < V; i++)
if (binary[i] == 1)
{
high = i;
count++;
}
return count;
}
int main()
{
int beauty[V][F] = {{7 , 5 , -21},
{23 , 21 , 5 },
{-5 , -4 , -4 },
{-24, 10 , -20},
{16 , 23 , 20 }};//定义美感数组
int best_beauty = 0, best_put = 0;//定义最大美感得分，和插花方案
int partial_sum[1<<V] = {0};
//枚举，从0到（2^V-1）
for (int i = 1; i < (1<<V); i++)
{
int binary[V] = {0};
ToBinary(i, binary);
int high;
int flowers = count1(binary, high);
if (flowers > F)//大于F非法方案，小于F局部方案，等于F完整插花方案
continue;
partial_sum[i] = partial_sum[i-(1<<high)] + beauty[high][flowers-1];
if (flowers == F && partial_sum[i] > best_beauty)
{
best_beauty = partial_sum[i];
best_put = i;
}
}
//输出答案
cout << "最大美感得分：" << best_beauty << endl;
cout << "插花方案：";
int best_binary[V] = {0};
ToBinary(best_put, best_binary);
for (int vase = 0, flower = 1; vase < V; vase++)
{
if (best_binary[vase] == 1)
{
cout << flower;
flower++;
}
else
cout << 0;
}
return 0;
}
6.3.5.1 动态规划解题，输出方案1
//橱窗插花，动态规划，输出方案1
#include <iostream>
using namespace std;
const int V = 5, F = 3;
//把一个数num表示为二进制数组binary[]
void ToBinary(int num, int binary[V])
{
for(int i = 0; i < V; i++)
{
binary[i] = num & 1; //按位与，取最低为
num = num >> 1; //右移一位
}
}
int main()
{
int beauty[V][F] = {{7 , 5 , -21},
{23 , 21 , 5 },
{-5 , -4 , -4 },
{-24, 10 , -20},
{16 , 23 , 20 }};//定义美感数组
int best_partial[V+1][F+1] = {{0}};
int best_put[V+1][F+1] = {{0}};
//m个花瓶插n朵花的局部最优解
for (int m = 1; m <= V; m++)
for (int n = 1; n <= m && n <= F; n++)
{
best_partial[m][n] = best_partial[m-1][n-1] + beauty[m-1][n-1];
best_put[m][n] = best_put[m-1][n-1] + (1<<(m-1));
if (n < m && best_partial[m][n] < best_partial[m-1][n])
{
best_partial[m][n] = best_partial[m-1][n];
best_put[m][n] = best_put[m-1][n];
}
cout << m << " "
<< n << " "
<< best_partial[m][n] << " "
<< best_put[m][n] << endl;
}
//输出答案
cout << "最大美感得分：" << best_partial[V][F] << endl;
cout << "插花方案：";
int best_binary[V] = {0};
ToBinary(best_put[V][F], best_binary);
for (int vase = 0, flower = 1; vase < V; vase++)
{
if (best_binary[vase] == 1)
{
cout << flower;
flower++;
}
else
cout << 0;
}
return 0;
}
6.3.5.2 动态规划，输出方案2
计算量和存储空间都得到优化，但输出为倒序
//橱窗插花，动态规划，输出方案2，回溯输出逆序
#include <iostream>
using namespace std;
const int V = 5, F = 3;
//把一个数num表示为二进制数组binary[]
int main()
{
int beauty[V][F] = {{7 , 5 , -21},
{23 , 21 , 5 },
{-5 , -4 , -4 },
{-24, 10 , -20},
{16 , 23 , 20 }};//定义美感数组
int best_partial[V+1][F+1] = {{0}};
bool best_put[V+1][F+1] = {{false}};
//m个花瓶插n朵花的局部最优解
for (int m = 1; m <= V; m++)
for (int n = 1; n <= m && n <= F; n++)
{
best_partial[m][n] = best_partial[m-1][n-1] + beauty[m-1][n-1];
best_put[m][n] = true;
if (n < m && best_partial[m][n] < best_partial[m-1][n])
{
best_partial[m][n] = best_partial[m-1][n];
best_put[m][n] = false;
}
cout << m << " "
<< n << " "
<< best_partial[m][n] << " "
<< best_put[m][n] << endl;
}
//输出答案
cout << "最大美感得分：" << best_partial[V][F] << endl;
cout << "插花方案：";
for (int m = V, n = F; m > 0; )
{
if (best_put[m][n])
{
cout << n;
m--;
n--;
}
else
{
cout << '0';
m--;
}
}
return 0;
}
6.3.5.3 动态规划，输出方案2，正序输出
思路是从第3朵花开始从第5个花瓶开始逆向插花：
所有的循环和递归方式不变，只是计算时将美感数组beauty[][]转置；
同样在输出时注意，逆序输出则第一个输出的是第1朵花；
//橱窗插花，动态规划,方案2，正序输出
#include <iostream>
using namespace std;
const int V = 5, F = 3;
int main()
{
int beauty[V][F] = {{7 , 5 , -21},
{23 , 21 , 5 },
{-5 , -4 , -4 },
{-24, 10 , -20},
{16 , 23 , 20 }};//定义美感数组
int best_partial[V+1][F+1] = {{0}};
bool best_put[V+1][F+1] = {{false}};
//m个花瓶插n朵花的局部最优解
for (int m = 1; m <= V; m++)
for (int n = 1; n <= m && n <= F; n++)
{
best_partial[m][n] = best_partial[m-1][n-1] + beauty[V-m][F-n];
best_put[m][n] = true;
if (n < m && best_partial[m][n] < best_partial[m-1][n])
{
best_partial[m][n] = best_partial[m-1][n];
best_put[m][n] = false;
}
cout << m << " "
<< n << " "
<< best_partial[m][n] << " "
<< best_put[m][n] << endl;
}
//输出答案
cout << "最大美感得分：" << best_partial[V][F] << endl;
cout << "插花方案：";
for (int m = V, n = F; m >= 1; )
if (best_put[m][n] == true)
{
cout << F-n+1;
m--;
n--;
}
else
{
cout << 0;
m--;
}
return 0;
}
6.4 最长公共子序列问题
递归初始值的确定，需要仔细分析，本节课中的做法非常巧妙；
存放长度的数组，可以用来进行比较，作为递归方向的判据；
存放方向的数组，作为递归方向的标志，可以作为输出时回溯的判定；
//最长公共子序列问题
#include <iostream>
#include <cstring>
using namespace std;
const int M = 10, N = 10;
int main()
{
char A[9] = "abcbdacb"; //要比较的字符串A
char B[6] = "bdcab"; //要比较的字符串B
int m = strlen(A); //字符串A的长度作为循环范围
int n = strlen(B); //字符串B的长度作为循环范围
int lcs[M][N]; //存放公共子序列的长度值
int decision[M][N]; //存放公共子序列计算时的取舍方向
enum //定义取舍方向的标志
{
I_J, //i和j相等，子序列长度加1，lcs[i][j]=1+lsc[i+1][j+1]
I_1, //i和j不相等，i+1比j公共子序列长度大，lsc[i][j]=lcs[i+1][j]
J_1 //i和j不相等，j+1比i公共子序列长度大，lsc[i][j]=lcs[i][j+1]
};
for (int i = 0; i < m+1; i++)
lcs[i][n] = 0; //初始化递归初值
for (int j = 0; j < n+1; j++)
lcs[m][j] = 0; //初始化递归初值
for (int i = m-1; i >= 0; i--) //递归运算
for (int j = n-1; j >= 0; j--)
if ( A[i] == B[j])
{
lcs[i][j] = 1 + lcs[i+1][j+1];
decision[i][j] = I_J;
}
else if ( lcs[i+1][j] > lcs[i][j+1])
{
lcs[i][j] = lcs[i+1][j];
decision[i][j] = I_1;
}
else
{
lcs[i][j] = lcs[i][j+1];
decision[i][j] = J_1;
}
for (int i = 0, j = 0; i < m && j < n; ) //回溯输出
switch (decision[i][j])
{
case I_J:
cout << A[i];
i++;
j++;
break;
case I_1:
i++;
break;
case J_1:
j++;
break;
}
return 0;
}
语法自测-试题答案
标签：binary,partial,int,put,学堂,num,第六章,程序设计,best 来源： https://blog.csdn.net/u014238501/article/details/120471329