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IOI '00 P5 - Post Office

Jun 8th, 2022
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  1. #include <iostream>
  2. #include <vector>
  3. #include <algorithm>
  4.  
  5. using namespace std;
  6.  
  7. const int INF = 1e9;
  8.  
  9. int main() {
  10.     int n, m;
  11.     cin >> n >> m;
  12.     vector<int> x(n);
  13.     for (int &a : x) cin >> a;
  14.     auto d = [&x](int i, int j) {
  15.         return abs(x[i]-x[j]);
  16.     };
  17.  
  18.     vector<vector<int>> dp(1+m, vector<int>(n, INF));
  19.     vector<vector<int>> prev(1+m, vector<int>(n, -1)); // previous post office used
  20.     // one post office
  21.     for (int i = 0; i < n; ++i) {
  22.         dp[1][i] = 0;
  23.         for (int j = 0; j < n; ++j) dp[1][i] += d(i, j);
  24.     }
  25.  
  26.     // more post offices
  27.     for (int k = 2; k <= m; ++k) { // number of post offices
  28.         for (int i = k-1; i < n; ++i) { // position of last post office
  29.             int firstCloser = i; // leftmost village closer to i than to j
  30.             int sumDistToI = 0; // sum of distances to i for all in range [firstCloser, i]
  31.             for (int j = i-1; j >= k-2; --j) { // position of previous post office
  32.                 // update firstCloser
  33.                 while (firstCloser && d(firstCloser-1, i) <= d(firstCloser-1, j)) {
  34.                     sumDistToI += d(--firstCloser, i);
  35.                 }
  36.  
  37.                 int deltaCost = 2*sumDistToI - (n-firstCloser) * d(i, j);
  38.                 int totalCost = dp[k-1][j] + deltaCost;
  39.                 if (totalCost < dp[k][i]) {
  40.                     dp[k][i] = totalCost;
  41.                     prev[k][i] = j;
  42.                 }
  43.             }
  44.         }
  45.     }
  46.  
  47.     int i = 0;
  48.     for (int j = 1; j < n; ++j) {
  49.         if (dp[m][j] < dp[m][i]) i = j;
  50.     }
  51.     cout << dp[m][i] << endl;
  52.     vector<int> v;
  53.     for (int k = m; k > 0; i=prev[k--][i]) v.push_back(i);
  54.     for (int p = m-1; p >= 0; --p) cout << x[v[p]] << ' ';
  55.     cout << endl;
  56.     return 0;
  57. }
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