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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.     // Problem: Split string into pieces to minimize cost
  15.     // Cost = (adjacent equal pairs) * c1 + (number of splits) * c2
  16.  
  17.     // pre[i] = number of adjacent equal pairs in s[0..i-1]
  18.     // Prefix sum of pairs (s[0],s[1]), (s[1],s[2]), ..., (s[i-2],s[i-1])
  19.     vector<int> pre(n + 1, 0);
  20.     for (int i = 1; i < n; i++) {
  21.         pre[i + 1] = pre[i];
  22.         if (s[i - 1] == s[i]) {
  23.             pre[i + 1]++;
  24.         }
  25.     }
  26.  
  27.     // dp[i] = min cost to optimally split s[0..i-1]
  28.     const long long INF = 1e18;
  29.     vector<long long> dp(n + 1, INF);
  30.     dp[0] = 0;  // empty string has cost 0
  31.  
  32.     // For each position i, try all possible ways to form the last piece
  33.     for (int i = 1; i <= n; i++) {
  34.         // Try making last piece from s[j..i-1]
  35.         for (int j = 0; j < i; j++) {
  36.             // Count adjacent equal pairs in substring s[j..i-1]
  37.             // Pairs are: (s[j],s[j+1]), (s[j+1],s[j+2]), ..., (s[i-2],s[i-1])
  38.             // These correspond to indices j+1, j+2, ..., i-1 in the prefix array
  39.             int cnt = pre[i] - pre[j + 1];
  40.  
  41.             // Cost of pairs in this piece
  42.             long long cost = (long long)cnt * c1;
  43.  
  44.             // Add split cost (c2) if this isn't the first piece
  45.             if (j > 0) cost += c2;
  46.  
  47.             // Update minimum cost to reach position i
  48.             dp[i] = min(dp[i], dp[j] + cost);
  49.         }
  50.     }
  51.  
  52.     cout << dp[n] << "\n";
  53.  
  54.     return 0;
  55. }
Success #stdin #stdout 0s 5276KB
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https://ideone.com/s0ALn7
language:
C++14 (gcc 8.3)
created:
3 hours ago
visibility:
public

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