# IOI '00 P5 - Post Office (2)

Jun 8th, 2022
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1. #include <bits/stdc++.h>
2.
3. using namespace std;
4.
5. const int INF = 1e9;
6.
7. int main() {
8.     int n, m;
9.     cin >> n >> m;
10.     vector<int> x(1+n);
11.     for (int i = 1; i <= n; ++i) cin >> x[i], x[i] += x[i-1];
12.
13.     auto A = [&x](int l, int r) {
14.         return (x[r]-x[r-(r-l+1)/2]) - (x[l+(r-l+1)/2-1]-x[l-1]);
15.     };
16.
17.     vector<vector<int>> dp(1+n, vector<int>(1+n, INF));
18.     vector<vector<int>> pos(1+n, vector<int>(2+n));
19.     for (int i = 1; i <= n; ++i) {
20.         dp[i][1] = A(1, i);
21.         pos[i][i+1] = i-1;
22.     }
23.     for (int i = 2; i <= n; ++i) {
24.         for (int j = i; j >= 2; --j) {
25.             for (int k = pos[i-1][j]; k <= pos[i][j+1]; ++k) {
26.                 dp[i][j] = min(dp[i][j], dp[k][j-1] + A(k+1, i));
27.             }
28.             for (int k = pos[i-1][j]; k <= pos[i][j+1]; ++k) {
29.                 if (dp[i][j] == dp[k][j-1] + A(k+1, i)) pos[i][j] = k;
30.             }
31.         }
32.     }
33.     cout << dp[n][m] << endl;
34.     vector<int> v;
35.     for (int i = n, j = m; j >= 1; i = pos[i][j--]) {
36.         int k = (i + pos[i][j]+1)/2;
37.         v.push_back(x[k]-x[k-1]);
38.     }
39.     for (int i = m-1; i >= 0; --i) cout << v[i] << ' ';
40.     cout << endl;
41.     return 0;
42. }