给定天数的最佳阅读列表
一个人决心在“k”天内完成这本书，但他从不想在这中间停下一章。找到章节的最佳分配，这样人们就不会总体上阅读过多/较少的页面。
例子：
Input: Number of Days to Finish book = 2
Number of pages in chapters = {10, 5, 5}
Output: Day 1: Chapter 1
Day 2: Chapters 2 and 3
Input: Number of Days to Finish book = 3
Number of pages in chapters = {8, 5, 6, 12}
Output: Day 1: Chapter 1
Day 2: Chapters 2 and 3
Day 2: Chapter 4
目标是最小化每天阅读的页面与平均页面数之间的差异总和。如果平均页数是非整数，则应四舍五入为最接近的整数。
在上面的示例 2 中，平均页数为 (8 + 5 + 6 + 12)/3 = 31/3，四舍五入为 10。因此，上面显示的输出的平均页数和每天的页数之间的差异是“ abs(8-10) + abs(5+6-10) + abs(12-10)”即5。值5是差值和的最佳值。
考虑上面的示例 2，其中一本书有 4 章，页数为 8、5、6 和 12。用户希望在 3 天内完成。上述场景的图形表示是，
在上图中，顶点表示章节，边 e(u, v) 表示从 'u' 到达 'v' 需要读取的页数。添加水槽节点以象征书的结尾。
首先，计算一天要阅读的平均页数（此处为 31/3，大约为 10）。新的边权重 e '(u, v) 将是平均差 |avg – e(u, v)|。上述问题的修改图将是，
感谢犰狳在评论中提出这个想法。
这个想法是从第 1 章开始，做一个 DFS 来找到边数为 'k' 的接收器。继续将访问过的顶点存储在一个数组中，比如“path[]”。如果到达目标顶点，且路径和小于最优路径，则更新最优分配optimal_path[]。请注意，该图是 DAG，因此无需处理 DFS 期间的循环。
下面，是相同的C++实现，邻接矩阵用于表示图。下面的程序主要有 4 个阶段。
1) 构造一个有向无环图。
2) 使用 DFS 找到最优路径。
3) 打印找到的最优路径。
C++
// C++ DFS solution to schedule chapters for reading in
// given days
# include
# include
# include
# include
using namespace std;
// Define total chapters in the book
// Number of days user can spend on reading
# define CHAPTERS 4
# define DAYS 3
# define NOLINK -1
// Array to store the final balanced schedule
int optimal_path[DAYS+1];
// Graph - Node chapter+1 is the sink described in the
// above graph
int DAG[CHAPTERS+1][CHAPTERS+1];
// Updates the optimal assignment with current assignment
void updateAssignment(int* path, int path_len);
// A DFS based recursive function to store the optimal path
// in path[] of size path_len. The variable sum stores sum of
// of all edges on current path. k is number of days spent so
// far.
void assignChapters(int u, int* path, int path_len, int sum, int k)
{
static int min = INT_MAX;
// Ignore the assignment which requires more than required days
if (k < 0)
return;
// Current assignment of chapters to days
path[path_len] = u;
path_len++;
// Update the optimal assignment if necessary
if (k == 0 && u == CHAPTERS)
{
if (sum < min)
{
updateAssignment(path, path_len);
min = sum;
}
}
// DFS - Depth First Search for sink
for (int v = u+1; v <= CHAPTERS; v++)
{
sum += DAG[u][v];
assignChapters(v, path, path_len, sum, k-1);
sum -= DAG[u][v];
}
}
// This function finds and prints optimal read list. It first creates a
// graph, then calls assignChapters().
void minAssignment(int pages[])
{
// 1) ............CONSTRUCT GRAPH.................
// Partial sum array construction S[i] = total pages
// till ith chapter
int avg_pages = 0, sum = 0, S[CHAPTERS+1], path[DAYS+1];
S[0] = 0;
for (int i = 0; i < CHAPTERS; i++)
{
sum += pages[i];
S[i+1] = sum;
}
// Average pages to be read in a day
avg_pages = round(sum/DAYS);
/* DAG construction vertices being chapter name &
* Edge weight being |avg_pages - pages in a chapter|
* Adjacency matrix representation */
for (int i = 0; i <= CHAPTERS; i++)
{
for (int j = 0; j <= CHAPTERS; j++)
{
if (j <= i)
DAG[i][j] = NOLINK;
else
{
sum = abs(avg_pages - (S[j] - S[i]));
DAG[i][j] = sum;
}
}
}
// 2) ............FIND OPTIMAL PATH................
assignChapters(0, path, 0, 0, DAYS);
// 3) ..PRINT OPTIMAL READ LIST USING OPTIMAL PATH....
cout << "Optimal Chapter Assignment :" << endl;
int ch;
for (int i = 0; i < DAYS; i++)
{
ch = optimal_path[i];
cout << "Day" << i+1 << ": " << ch << " ";
ch++;
while ( (i < DAYS-1 && ch < optimal_path[i+1]) ||
(i == DAYS-1 && ch <= CHAPTERS))
{
cout << ch << " ";
ch++;
}
cout << endl;
}
}
// This function updates optimal_path[]
void updateAssignment(int* path, int path_len)
{
for (int i = 0; i < path_len; i++)
optimal_path[i] = path[i] + 1;
}
// Driver program to test the schedule
int main(void)
{
int pages[CHAPTERS] = {7, 5, 6, 12};
// Get read list for given days
minAssignment(pages);
return 0;
} Python3
# Python3 DFS solution to schedule chapters
# for reading in given days
# A DFS based recursive function to store
# the optimal path in path[] of size path_len.
# The variable Sum stores Sum of all edges on
# current path. k is number of days spent so far.
def assignChapters(u, path, path_len, Sum, k):
global CHAPTERS, DAYS, NOLINK, optical_path, DAG, Min
# Ignore the assignment which requires
# more than required days
if (k < 0):
return
# Current assignment of chapters to days
path[path_len] = u
path_len += 1
# Update the optimal assignment if necessary
if (k == 0 and u == CHAPTERS):
if (Sum < Min):
updateAssignment(path, path_len)
Min = Sum
# DFS - Depth First Search for sink
for v in range(u + 1, CHAPTERS + 1):
Sum += DAG[u][v]
assignChapters(v, path, path_len, Sum, k - 1)
Sum -= DAG[u][v]
# This function finds and prints
# optimal read list. It first creates a
# graph, then calls assignChapters().
def MinAssignment(pages):
global CHAPTERS, DAYS, NOLINK, optical_path, DAG, MIN
# 1) ............CONSTRUCT GRAPH.................
# Partial Sum array construction S[i] = total pages
# till ith chapter
avg_pages = 0
Sum = 0
S = [None] * (CHAPTERS + 1)
path = [None] * (DAYS + 1)
S[0] = 0
for i in range(CHAPTERS):
Sum += pages[i]
S[i + 1] = Sum
# Average pages to be read in a day
avg_pages = round(Sum/DAYS)
# DAG construction vertices being chapter name &
# Edge weight being |avg_pages - pages in a chapter|
# Adjacency matrix representation
for i in range(CHAPTERS + 1):
for j in range(CHAPTERS + 1):
if (j <= i):
DAG[i][j] = NOLINK
else:
Sum = abs(avg_pages - (S[j] - S[i]))
DAG[i][j] = Sum
# 2) ............FIND OPTIMAL PATH................
assignChapters(0, path, 0, 0, DAYS)
# 3) ..PROPTIMAL READ LIST USING OPTIMAL PATH....
print("Optimal Chapter Assignment :")
ch = None
for i in range(DAYS):
ch = optimal_path[i]
print("Day", i + 1, ": ", ch, end = " ")
ch += 1
while ((i < DAYS - 1 and ch < optimal_path[i + 1]) or
(i == DAYS - 1 and ch <= CHAPTERS)):
print(ch, end = " ")
ch += 1
print()
# This function updates optimal_path[]
def updateAssignment(path, path_len):
for i in range(path_len):
optimal_path[i] = path[i] + 1
# Driver Code
# Define total chapters in the book
# Number of days user can spend on reading
CHAPTERS = 4
DAYS = 3
NOLINK = -1
# Array to store the final balanced schedule
optimal_path = [None] * (DAYS + 1)
# Graph - Node chapter+1 is the sink
# described in the above graph
DAG = [[None] * (CHAPTERS + 1)
for i in range(CHAPTERS + 1)]
Min = 999999999999
pages = [7, 5, 6, 12]
# Get read list for given days
MinAssignment(pages)
# This code is contributed by PranchalK输出：
Optimal Chapter Assignment :
Day1: 1
Day2: 2 3
Day3: 4如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。