给定一个数组arr[]由N 个正整数组成，表示N 个项目的重量和一个正整数D ，任务是找到一艘船的最小承重能力（比如K ）在D天内运送所有重量，使得订单船上装载的重量按arr[] 中数组元素的顺序排列，船舶每天装载的重量总量为K 。
例子：
Input: arr[] = {1, 2, 1}, D = 2
Output: 3
Explanation:
Consider the minimum weight required by the boat as 3, then below is the order of weights such all the weight can be shipped within D(= 2) days:
Day 1: Ship the weights of values 1 and 2 on the first day as the sum of weights 1 + 2 = 3(<= 3).
Day 2: Ship the weights of value 1 on the second day as the sum of weights 1(<= 3).
Considering the minimum weight amount as 3, ships all the weight within D(= 2) days. Therefore, print 3.
Input: arr[] = {9, 8, 10}, D = 3
Output: 10
方法：给定的问题可以通过使用贪心技术和二分搜索来解决。问题的单调性可以看出，如果所有包裹都可以在D天内以容量K成功发货，那么它们肯定可以以任何大于K 的容量发货。请按照以下步骤解决问题：
- 初始化一个变量，比如ans为-1来存储所需的船的最小容量。
- 用给定数组中的最大元素和数组的总和分别初始化两个变量，比如s和e ，这表示搜索空间的下限和上限。
- 迭代直到s的值小于或等于e ，并执行以下步骤：
- 初始化一个变量，比如mid为(s + e)/2 。
- 检查是否可以在允许的最大容量为mid 的D天内运送所有包裹。如果发现为true ，则将ans的值更新为mid ，将e的值更新为(mid – 1) 。
- 否则，将s的值更新为(mid + 1) 。
- 完成以上步骤后，打印ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if the weights
// can be delivered in D days or not
bool isValid(int weight[], int n,
int D, int mx)
{
// Stores the count of days required
// to ship all the weights if the
// maximum capacity is mx
int st = 1;
int sum = 0;
// Traverse all the weights
for (int i = 0; i < n; i++) {
sum += weight[i];
// If total weight is more than
// the maximum capacity
if (sum > mx) {
st++;
sum = weight[i];
}
// If days are more than D,
// then return false
if (st > D)
return false;
}
// Return true for the days < D
return true;
}
// Function to find the least weight
// capacity of a boat to ship all the
// weights within D days
void shipWithinDays(int weight[], int D,
int n)
{
// Stores the total weights to
// be shipped
int sum = 0;
// Find the sum of weights
for (int i = 0; i < n; i++)
sum += weight[i];
// Stores the maximum weight in the
// array that has to be shipped
int s = weight[0];
for (int i = 1; i < n; i++) {
s = max(s, weight[i]);
}
// Store the ending value for
// the search space
int e = sum;
// Store the required result
int res = -1;
// Perform binary search
while (s <= e) {
// Store the middle value
int mid = s + (e - s) / 2;
// If mid can be shipped, then
// update the result and end
// value of the search space
if (isValid(weight, n, D, mid)) {
res = mid;
e = mid - 1;
}
// Search for minimum value
// in the right part
else
s = mid + 1;
}
// Print the result
cout << res;
}
// Driver Code
int main()
{
int weight[] = { 9, 8, 10 };
int D = 3;
int N = sizeof(weight) / sizeof(weight[0]);
shipWithinDays(weight, D, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to check if the weights
// can be delivered in D days or not
static boolean isValid(int[] weight, int n,
int D, int mx)
{
// Stores the count of days required
// to ship all the weights if the
// maximum capacity is mx
int st = 1;
int sum = 0;
// Traverse all the weights
for(int i = 0; i < n; i++)
{
sum += weight[i];
// If total weight is more than
// the maximum capacity
if (sum > mx)
{
st++;
sum = weight[i];
}
// If days are more than D,
// then return false
if (st > D)
return false;
}
// Return true for the days < D
return true;
}
// Function to find the least weight
// capacity of a boat to ship all the
// weights within D days
static void shipWithinDays(int[] weight, int D, int n)
{
// Stores the total weights to
// be shipped
int sum = 0;
// Find the sum of weights
for(int i = 0; i < n; i++)
sum += weight[i];
// Stores the maximum weight in the
// array that has to be shipped
int s = weight[0];
for(int i = 1; i < n; i++)
{
s = Math.max(s, weight[i]);
}
// Store the ending value for
// the search space
int e = sum;
// Store the required result
int res = -1;
// Perform binary search
while (s <= e)
{
// Store the middle value
int mid = s + (e - s) / 2;
// If mid can be shipped, then
// update the result and end
// value of the search space
if (isValid(weight, n, D, mid))
{
res = mid;
e = mid - 1;
}
// Search for minimum value
// in the right part
else
s = mid + 1;
}
// Print the result
System.out.println(res);
}
// Driver Code
public static void main(String[] args)
{
int[] weight = { 9, 8, 10 };
int D = 3;
int N = weight.length;
shipWithinDays(weight, D, N);
}
}
// This code is contributed by Dharanendra L V.Python3
# Python3 program for the above approach
# Function to check if the weights
# can be delivered in D days or not
def isValid(weight, n, D, mx):
# Stores the count of days required
# to ship all the weights if the
# maximum capacity is mx
st = 1
sum = 0
# Traverse all the weights
for i in range(n):
sum += weight[i]
# If total weight is more than
# the maximum capacity
if (sum > mx):
st += 1
sum = weight[i]
# If days are more than D,
# then return false
if (st > D):
return False
# Return true for the days < D
return True
# Function to find the least weight
# capacity of a boat to ship all the
# weights within D days
def shipWithinDays(weight, D, n):
# Stores the total weights to
# be shipped
sum = 0
# Find the sum of weights
for i in range(n):
sum += weight[i]
# Stores the maximum weight in the
# array that has to be shipped
s = weight[0]
for i in range(1, n):
s = max(s, weight[i])
# Store the ending value for
# the search space
e = sum
# Store the required result
res = -1
# Perform binary search
while (s <= e):
# Store the middle value
mid = s + (e - s) // 2
# If mid can be shipped, then
# update the result and end
# value of the search space
if (isValid(weight, n, D, mid)):
res = mid
e = mid - 1
# Search for minimum value
# in the right part
else:
s = mid + 1
# Print the result
print(res)
# Driver Code
if __name__ == '__main__':
weight = [ 9, 8, 10 ]
D = 3
N = len(weight)
shipWithinDays(weight, D, N)
# This code is contributed by ipg2016107C#
// C# program for the above approach
using System;
class GFG{
// Function to check if the weights
// can be delivered in D days or not
static bool isValid(int[] weight, int n,
int D, int mx)
{
// Stores the count of days required
// to ship all the weights if the
// maximum capacity is mx
int st = 1;
int sum = 0;
// Traverse all the weights
for(int i = 0; i < n; i++)
{
sum += weight[i];
// If total weight is more than
// the maximum capacity
if (sum > mx)
{
st++;
sum = weight[i];
}
// If days are more than D,
// then return false
if (st > D)
return false;
}
// Return true for the days < D
return true;
}
// Function to find the least weight
// capacity of a boat to ship all the
// weights within D days
static void shipWithinDays(int[] weight, int D, int n)
{
// Stores the total weights to
// be shipped
int sum = 0;
// Find the sum of weights
for(int i = 0; i < n; i++)
sum += weight[i];
// Stores the maximum weight in the
// array that has to be shipped
int s = weight[0];
for(int i = 1; i < n; i++)
{
s = Math.Max(s, weight[i]);
}
// Store the ending value for
// the search space
int e = sum;
// Store the required result
int res = -1;
// Perform binary search
while (s <= e)
{
// Store the middle value
int mid = s + (e - s) / 2;
// If mid can be shipped, then
// update the result and end
// value of the search space
if (isValid(weight, n, D, mid))
{
res = mid;
e = mid - 1;
}
// Search for minimum value
// in the right part
else
s = mid + 1;
}
// Print the result
Console.WriteLine(res);
}
// Driver Code
public static void Main()
{
int[] weight = { 9, 8, 10 };
int D = 3;
int N = weight.Length;
shipWithinDays(weight, D, N);
}
}
// This code is contributed by ukaspJavascript
输出：
10时间复杂度： O(N*log(S – M))，其中S是数组元素的总和， M是数组的最大元素。
辅助空间： O(1)
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