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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.     // Example: "aabbc" -> pre = [0, 0, 1, 1, 2, 2]
  19.     vector<int> pre(n + 1, 0);
  20.     for (int i = 0; i < n - 1; i++) {
  21.         pre[i + 1] = pre[i];
  22.         if (s[i] == s[i + 1]) {
  23.             pre[i + 1]++;
  24.         }
  25.     }
  26.     pre[n] = pre[n - 1];
  27.  
  28.     // dp[i] = min cost to optimally split s[0..i-1]
  29.     const long long INF = 1e18;
  30.     vector<long long> dp(n + 1, INF);
  31.     dp[0] = 0;  // empty string has cost 0
  32.  
  33.     // For each position i, try all possible ways to form the last piece
  34.     for (int i = 1; i <= n; i++) {
  35.         // Try making last piece from s[j..i-1]
  36.         for (int j = 0; j < i; j++) {
  37.             // Count adjacent equal pairs in range [j, i)
  38.             int cnt = pre[i] - pre[j];
  39.  
  40.             // Cost of pairs in this piece
  41.             long long cost = (long long)cnt * c1;
  42.  
  43.             // Add split cost (c2) if this isn't the first piece
  44.             if (j > 0) cost += c2;
  45.  
  46.             // Update minimum cost to reach position i
  47.             dp[i] = min(dp[i], dp[j] + cost);
  48.         }
  49.     }
  50.  
  51.     cout << dp[n] << "\n";
  52.  
  53.     return 0;
  54. }
Success #stdin #stdout 0.01s 5280KB
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https://ideone.com/Jj8WFF
language:
C++14 (gcc 8.3)
created:
4 hours ago
visibility:
public

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