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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. int main() {
  5.     ios::sync_with_stdio(false);
  6.     cin.tie(nullptr);
  7.  
  8.     long long c1, c2;
  9.     string s;
  10.     cin >> c1 >> c2 >> s;
  11.  
  12.     int n = s.size();
  13.  
  14.     // pre[i] = number of adjacent equal pairs in first i characters
  15.     vector<int> pre(n + 1, 0);
  16.     for (int i = 0; i < n - 1; i++) {
  17.         pre[i + 1] = pre[i];
  18.         if (s[i] == s[i + 1]) {
  19.             pre[i + 1]++;
  20.         }
  21.     }
  22.     pre[n] = pre[n - 1];
  23.  
  24.     // dp[i] = min cost to handle first i characters
  25.     const long long INF = 1e18;
  26.     vector<long long> dp(n + 1, INF);
  27.     dp[0] = 0;
  28.  
  29.     // Try every position
  30.     for (int i = 1; i <= n; i++) {
  31.         // Try every possible start for last piece
  32.         for (int j = 0; j < i; j++) {
  33.             // Count pairs in range [j, i)
  34.             int cnt = pre[i] - pre[j];
  35.  
  36.             // Cost of pairs in this piece
  37.             long long cost = (long long)cnt * c1;
  38.  
  39.             // Add split cost if not first piece
  40.             if (j > 0) cost += c2;
  41.  
  42.             // Update answer
  43.             dp[i] = min(dp[i], dp[j] + cost);
  44.         }
  45.     }
  46.  
  47.     cout << dp[n] << "\n";
  48.  
  49.     return 0;
  50. }
Success #stdin #stdout 0.01s 5272KB
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https://ideone.com/3fkVvW
language:
C++14 (gcc 8.3)
created:
4 hours ago
visibility:
public

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