最大限度地降低运营成本以平衡塔的高度
给定 n (n <=10000) 塔的高度作为数组 h[];我们需要通过在塔中添加或移除块来使每个塔都达到相同的高度。在不同的塔中，每次添加或移除操作都会花费不同的价值。目标是最小化这个成本。
例子：
Input : Tower heights h[] = {1, 2, 3}
Costs of operations cost[] = {10, 100, 1000}
Output : 120
The heights can be equalized by either "Removing
one block from 3 and adding one in 1" or "Adding
two blocks in 1 and adding one in 2". Since the
cost of operation in tower 3 is 1000, the first
process would yield 1010 while the second one
yields 120. Since the second process yields the
lowest cost of operation, it is the required
output.
Input : h[] = {7,1,5 };
cost[] = { 1, 1, 1 };
Output : 6
abs(7-5) + abs(1-5) + abs(5-5) = 6 (taking 5
as final height)
Input : h[] = {2, 4, 8}
cost[] = {10, 100, 1000}
Output : 460朴素方法：一种朴素的方法是找到每个可能的操作集的成本，然后找到最小的操作成本。在最坏的情况下，这将给出 O(n 2 )。
最佳方法：这里最好的方法似乎是二进制搜索。我们存储最小高度并在每一步中将结果高度更改为 2 倍。
C++
// C++ program to minimize the cost of operation
// to bring all towers to same height.
#include
using namespace std;
// Returns the cost of entire operation in bringing
// the height of all towers to equal height eq_h
long int costOfOperation(int n, int h[], int cost[],
int eq_h)
{
// Initialize initial cost to 0
long int c = 0;
// Adding cost for each tower
for (int i = 0; i < n; i++)
c += abs(h[i] - eq_h) * cost[i];
return c;
}
// Return the minimum possible cost of operation
// to bring all the towers of different height
// in height[0..n-1] to same height.
long long int Bsearch(int n, int h[], int cost[])
{
long int max_h = *max_element(h, h + n);
long int ans = LONG_MAX;
// Do binary search for minimum cost
long int high = 1 + max_h;
long int low = 0;
while (high > low) {
int mid = (low + high) >> 1;
// Cost below mid
long int bm = (mid > 0) ?
costOfOperation(n, h, cost, mid - 1) :
LONG_MAX;
// Cost at mid
long int m = costOfOperation(n, h, cost, mid);
// Cost above mid
long int am = costOfOperation(n, h, cost, mid + 1);
// ans should hold the minimum cost of operation
ans = min(ans, m);
// Search lower cost
if (bm <= m)
high = mid;
// Search higher cost
else if (am <= m)
low = mid + 1;
// If am > m and bm > m
// then ans is m
else
return m;
}
return ans;
}
// Driver code
int main()
{
int h[] = { 1, 2, 3 };
int cost[] = { 10, 100, 1000 };
int n = sizeof(h)/sizeof(h[0]);
cout << Bsearch(n, h, cost);
return 0;
} Java
// Java program to minimize the cost of operation
// to bring all towers to same height.
import java.util.Arrays;
public class MinCostOp_Mintowerheight {
// Returns the cost of entire operation in bringing
// the height of all towers to equal height eq_h
static long costOfOperation(int n, int h[], int cost[],
int eq_h)
{
// Initialize initial cost to 0
long c = 0;
// Adding cost for each tower
for (int i = 0; i < n; i++)
c += Math.abs(h[i] - eq_h) * cost[i];
return c;
}
// Return the minimum possible cost of operation
// to bring all the towers of different height
// in height[0..n-1] to same height.
static long Bsearch(int n, int h[], int cost[])
{
int max_h = Arrays.stream(h).max().getAsInt();
long ans = Long.MAX_VALUE;
// Do binary search for minimum cost
long high = 1 + max_h;
long low = 0;
while (high > low) {
int mid = (int) ((low + high) >> 1);
// Cost below mid
long bm = (mid > 0) ?
costOfOperation(n, h, cost, mid - 1) :
Long.MAX_VALUE;
// Cost at mid
long m = costOfOperation(n, h, cost, mid);
// Cost above mid
long am = costOfOperation(n, h, cost, mid + 1);
// Break if the answer becomes equal to
// minimum cost m
if (ans == m)
break;
// ans should hold the minimum cost of operation
ans = Long.min(ans, m);
// Search lower cost
if (bm <= m)
high = mid;
// Search higher cost
else if (am <= m)
low = mid + 1;
// If am > m and bm > m
// then ans is m
else
return m;
}
return ans;
}
// Driver code
public static void main(String args[])
{
int h[] = { 1, 2, 3 };
int cost[] = { 10, 100, 1000 };
int n = h.length;
System.out.println(Bsearch(n, h, cost));
}
}
// This code is contributed by Sumit GhoshPython3
# Python 3 program to minimize the cost of
# operation to bring all towers to same height.
import sys
# Returns the cost of entire operation in
# bringing the height of all towers to
# equal height eq_h
def costOfOperation(n, h,cost, eq_h):
# Initialize initial cost to 0
c = 0
# Adding cost for each tower
for i in range(0, n, 1):
c += abs(h[i] - eq_h) * cost[i]
return c
# Return the minimum possible cost of
# operation to bring all the towers of
# different height in height[0..n-1]
# to same height.
def Bsearch(n, h, cost):
max_h = h[0]
for i in range(len(h)):
if(h[i] > max_h):
max_h = h[i]
ans = sys.maxsize
# Do binary search for minimum cost
high = 1 + max_h
low = 0
while (high > low):
mid = (low + high) >> 1
# Cost below mid
if(mid > 0):
bm = costOfOperation(n, h, cost, mid - 1)
else:
bm = sys.maxsize
# Cost at mid
m = costOfOperation(n, h, cost, mid)
# Cost above mid
am = costOfOperation(n, h, cost, mid + 1)
# Break if the answer becomes equal
# to minimum cost m
if (ans == m):
break
# ans should hold the minimum cost
# of operation
ans = min(ans, m)
# Search lower cost
if (bm <= m):
high = mid
# Search higher cost
else if(am <= m):
low = mid + 1
# If am > m and bm > m
# then ans is m
else :
return m;
return ans
# Driver code
if __name__ == '__main__':
h = [1, 2, 3]
cost = [10, 100, 1000]
n = len(h)
print(Bsearch(n, h, cost))
# This code is contributed by
# Sahil_shelangiaC#
// C# program to minimize the
// cost of operation to bring
// all towers to same height.
using System;
using System.Linq;
public class MinCostOp_Mintowerheight
{
// Returns the cost of entire
// operation in bringing the height
// of all towers to equal height eq_h
static long costOfOperation(int n, int []h,
int []cost, int eq_h)
{
// Initialize initial cost to 0
long c = 0;
// Adding cost for each tower
for (int i = 0; i < n; i++)
c += Math.Abs(h[i] - eq_h) * cost[i];
return c;
}
// Return the minimum possible
// cost of operation to bring
// all the towers of different height
// in height[0..n-1] to same height.
static long Bsearch(int n, int []h, int []cost)
{
int max_h = h.Max();
long ans = long.MaxValue;
// Do binary search for minimum cost
long high = 1 + max_h;
long low = 0;
while (high > low)
{
int mid = (int) ((low + high) >> 1);
// Cost below mid
long bm = (mid > 0) ?
costOfOperation(n, h, cost, mid - 1) :
long.MaxValue;
// Cost at mid
long m = costOfOperation(n, h, cost, mid);
// Cost above mid
long am = costOfOperation(n, h, cost, mid + 1);
// Break if the answer becomes
// equal to minimum cost m
if (ans == m)
break;
// ans should hold the minimum
// cost of operation
ans = Math.Min(ans, m);
// Search lower cost
if (bm <= m)
high = mid;
// Search higher cost
else if (am <= m)
low = mid + 1;
// If am > m and bm > m
// then ans is m
else
return m;
}
return ans;
}
// Driver code
public static void Main(String []args)
{
int []h = { 1, 2, 3 };
int []cost = { 10, 100, 1000 };
int n = h.Length;
Console.WriteLine(Bsearch(n, h, cost));
}
}
// This code contributed by Rajput-JiPHP
$low)
{
$mid = ($low + $high) >> 1;
// Cost below mid
$bm = ($mid > 0) ?
costOfOperation($n, $h, $cost, $mid - 1) :
PHP_INT_MAX;
// Cost at mid
$m = costOfOperation($n, $h, $cost, $mid);
// Cost above mid
$am = costOfOperation($n, $h, $cost, $mid + 1);
// Break if the answer becomes equal to
// minimum cost m
if ($ans == $m)
break;
// ans should hold the minimum
// cost of operation
$ans = min($ans, $m);
// Search lower cost
if ($bm <= $m)
$high = $mid;
// Search higher cost
else if ($am <= $m)
$low = $mid + 1;
// If am > m and bm > m
// then ans is m
else
return $m;
}
return $ans;
}
// Driver code
$h = array( 1, 2, 3 );
$cost = array( 10, 100, 1000 );
$n = count($h);
echo Bsearch($n, $h, $cost);
// This code is contributed by mits
?>Javascript
输出：
120