Algorithm
Problem link- https://www.spoj.com/problems/KOPC12A/
KOPC12A - K12 - Building Construction
Given N buildings of height h1, h2, h3 ... hn, the objective is to make every building has equal height. This can be done by removing bricks from a building or adding some bricks to a building. Removing a brick or adding a brick is done at certain cost which will be given along with the heights of the buildings. Find the minimal cost at which you can make the buildings look beautiful by re-constructing the buildings such that the N buildings satisfy
h1 = h2 = h3 = ... = hn = k (k can be any number).
For convenience, all buildings are considered to be vertical piles of bricks, which are of same dimensions.
Input
The first line of input contains an integer T which denotes number of test cases .This will be followed by 3 * T lines, 3 lines per test case. The first line of each test case contains an integer n and the second line contains n integers which denotes the heights of the buildings [h1, h2, h3 ... hn] and the third line contains n integers [c1, c2, c3 ... cn] which denotes the cost of adding or removing one unit of brick from the corresponding building.
T <= 15; n <= 10000; 0 <= Hi <= 10000; 0 <= Ci <= 10000;
Output
The output must contain T lines each line corresponding to a testcase.
Example
Input: 1 3 1 2 3 10 100 1000 Output: 120
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
#define MOD (ll)1000000007
#define pb push_back
#define EPS 1e-9
#define FOR(i, n) for(int i = 0;i < n; i++)
#define pi(a) printf("%d\n", a)
#define all(c) c.begin(), c.end()
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}
int n;
ll height[10005];
ll cost[10005];
int isIncreasing(ll mid){
//find cost at mid
//find cost at (mid+1)
ll ans1 = 0;
ll ans2 = 0;
for(int i = 0;i < n; i++){
ans1 = ans1 + (cost[i]*(abs(height[i]-mid)));
ans2 = ans2 + (cost[i]*(abs(height[i]-(mid+1))));
}
if(ans2 >= ans1)
return 1;
else
return 0;
}
int main(){
int t;
si(t);
while(t--){
// int n;
si(n);
// cout<<n<<endl;
// int height[n];
FOR(i, n) cin>>height[i];
// int cost[n];
FOR(i, n) cin>>cost[i];
ll minn = LONG_MAX, maxx = LONG_MIN;
for(int i = 0;i < n; i++){
minn = min(minn, height[i]);
maxx = max(maxx, height[i]);
}
// cout<<minn<<' '<<maxx<<endl;
ll lo = minn-1, hi = maxx;
while(hi-lo > 1){
ll mid = (hi+lo)/2;
// cout<<mid<<endl;
if(isIncreasing(mid)){
hi = mid;
}else{
lo = mid;
}
}
ll ans1 = 0;
for(int i = 0;i < n; i++){
ans1 = ans1 + (cost[i]*(abs(height[i]-hi)));
}
cout<<ans1<<endl;
}
return 0;
}Input
3
1 2 3
10 100 1000
Output
Demonstration
SPOJ Solution-Building Construction-Solution in C, C++, Java, Python