Mining
Problem Statement :
There are n gold mines along a river, and each mine i produces wi tons of gold. In order to collect the mined gold, we want to redistribute and consolidate it amongst exactly k mines where it can be picked up by trucks. We do this according to the following rules: You can move gold between any pair of mines (i.e., i and j, where 1 <= i <j <= n). All the gold at some pickup mine i must either stay at mine i or be completely moved to some other mine, j. Move w tons of gold between the mine at location xi and the mine at location xj at a cost of |xi-xj|*w. Given n, k, and the amount of gold produced at each mine, find and print the minimum cost of consolidating the gold into k pickup locations according to the above conditions. Input Format The first line contains two space-separated integers describing the respective values of n (the number of mines) and k (the number of pickup locations). Each line i of the n subsequent lines contains two space-separated integers describing the respective values of xi (the mine's distance from the mouth of the river) and wi (the amount of gold produced in tons) for mine i. Note: It is guaranteed that the mines are will be given in order of ascending location. Constraints 1 <= k < n <= 5000 1 <= wi,xi <= 10^6 Output Format Print a single line with the minimum cost of consolidating the mined gold amongst k different pickup sites according to the rules stated above.
Solution :
Solution in C :
In C++ :
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0);
for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
const int M=5010;
ll wt[M],dp[M][M],w[M][M],sw[M],scw[M];
int s[M][M],cord[M];
int n,p;
ll calc(int i, int j,int k) {
assert(i<=k&&k<=j);
return cord[k]*(sw[k]-sw[i-1])-(scw[k]-scw[i-1])+
scw[j]-scw[k-1]-cord[k]*(sw[j]-sw[k-1]);
}
int main() {
scanf("%d%d",&n,&p);
rep(i,1,n+1) {
scanf("%d%lld",cord+i,wt+i);
}
rep(i,1,n+1) sw[i]=sw[i-1]+wt[i],scw[i]=scw[i-1]+cord[i]*wt[i];
rep(i,1,n+1) {
int p=i;
rep(j,i+1,n+1) {
while (p<j&&calc(i,j,p)>calc(i,j,p+1)) ++p;
w[i][j]=calc(i,j,p);
}
}
rep(i,1,n+1) dp[1][i]=w[1][i],s[i][i]=i-1;
rep(i,2,p+1) {
int j=n;
dp[i][n]=1ll<<60;
for (int k = s[i - 1][n]; k <= n - 1; k++) {
ll temp = dp[i - 1][k] + w[k + 1][n];
if (temp < dp[i][n]) dp[i][n] = temp,s[i][n] = k;
}
for (int j = n - 1; j >= i + 1; j--) {
dp[i][j] = 1ll<<60;
for (int k = s[i - 1][j]; k <= s[i][j + 1]; k++) {
ll temp = dp[i - 1][k] + w[k + 1][j];
if (temp < dp[i][j]) dp[i][j] = temp,s[i][j] = k;
}
}
}
printf("%lld\n",dp[p][n]);
}
In Java :
import java.io.File;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
// Scanner scn = null;
// try {
// scn = new Scanner(new File("input.txt"));
// } catch (Exception ex) {
// ex.printStackTrace();
// }
int n = scn.nextInt();
int m = scn.nextInt();
int[] x = new int[n];
long[] w = new long[n];
for (int i = 0; i < n; ++i) {
x[i] = scn.nextInt();
w[i] = scn.nextLong();
}
long f[][] = new long[n][m + 1];
int b[][] = new int[n][m + 1];
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= m; ++j) {
f[i][j] = Long.MAX_VALUE / 2;
}
}
f[0][1] = 0;
long s = w[0];
for (int i = 1; i < n; ++i) {
f[i][1] = f[i - 1][1] + s * (x[i] - x[i - 1]);
s += w[i];
}
for (int j = 2; j <= m; ++j) {
for (int i = j - 1; i < n; ++i) {
int l = i;
s = w[i];
long t = w[i];
long cost = 0;
for (int k = i - 1; k >= j - 2; --k) {
cost += (s - t) * (x[k + 1] - x[k]);
s += w[k];
while (l - 1 >= k && x[i] - x[l - 1] <= x[l - 1] - x[k]) {
l--;
t += w[l];
cost += w[l] * (x[i] - x[l]) - w[l] * (x[l] - x[k]);
}
if (f[i][j] > f[k][j - 1] + cost) {
f[i][j] = f[k][j - 1] + cost;
b[i][j] = k;
}
if (k < b[i - 1][j]) {
break;
}
// System.out.println("current cost = " + (f[k][j - 1] + cost));
}
// System.out.println("+++");
}
}
long ret = f[n - 1][m];
// System.out.println("ret = " + ret);
s = w[n - 1];
long cost = 0;
for (int i = n - 2; i >= m - 1; --i) {
// System.out.println("i = " + i);
// System.out.println("fim = " + f[i][m]);
cost += s * (x[i + 1] - x[i]);
// System.out.println("cost = " + cost);
ret = Math.min(ret, f[i][m] + cost);
s += w[i];
// System.out.println("ret = " + ret);
// System.out.println("+++");
}
System.out.println(ret);
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
int x[5000];
int w[5000];
int64_t val[5000][5000];
int64_t a[5000], b[5000];
int64_t mining()
{
int n, k, i;
scanf("%d %d", &n, &k);
for(i = 0; i < n; ++ i)
scanf("%d %d", &x[i], &w[i]);
for(i = 0; i < n; ++i)
{
int64_t left = 0, right = 0, acc = 0;
int s, j;
for(j = i + 1, s = i; j < n; ++ j)
{
acc += w[j] * (int64_t)(x[j] - x[s]);
right += w[j];
while(s < j && left + w[s] < right)
{
acc += (left + w[s] - right) * (int64_t)(x[s + 1] - x[s]);
left += w[s];
right -= w[s + 1];
++ s;
}
val[j][i] = acc;
}
}
/* memcpy(a, val[0], n * sizeof(int64_t)); */
for(i = 0; i < n; ++ i)
a[i] = val[i][0];
if(n * (int64_t) n * (int64_t) k < 1000000000)
{
for(; 1 < k; --k)
for(i = n - 1; -1 < i; --i)
{
int s;
a[i] = val[i][0];
for(s = 1; s < i + 1; ++ s)
{
const int64_t c = a[s - 1] + val[i][s];
if(c < a[i]) a[i] = c;
}
}
return a[n - 1];
}
for(; 1 < k; -- k)
{
int idx = 0;
memcpy(b, a, n * sizeof(int64_t));
for(i = 0; i < n; ++ i)
{
a[i] = (idx ? b[idx - 1] : 0) + val[i][idx];
for(; idx < i && b[idx] + val[i][idx + 1] < a[i]; ++ idx)
a[i] = b[idx] + val[i][idx + 1];
{
int s = i;
for(s = i; idx < s && i < s + 50; -- s)
{
const int64_t v = (s ? b[s - 1] : 0) + val[i][s];
if(v < a[i]) a[i] = v;
}
}
}
}
return a[n - 1];
}
int main()
{
printf("%lld", (long long) mining());
return EXIT_SUCCESS;
}
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