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Feb 20th, 2018
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  1. // 0,1,2 - цвет - никакой, 1-ой компании, 2-ой компании
  2.  
  3. dp[0][0] = 0;
  4. dp[1][0] = dp[2][0] = 1;
  5.  
  6. for (int i = 0; i < k-1; i++) {
  7.  
  8. //next = TODO
  9.  
  10. dp[0][next] = max(
  11. dp[0][prev] + dist1[0][prev] + dist2[0][prev] - 2,
  12. dp[1][prev] + dist1[0][prev] + dist2[0][prev] - 2,
  13. dp[2][prev] + dist1[0][prev] + dist2[0][prev] - 2
  14. );
  15.  
  16. dp[1][next] = max(
  17. dp[0][prev] + f(dist1[0][prev], 0, 0) + f(dist2[0][prev], 0, 0) - 1,
  18. dp[1][prev] + f(dist1[0][prev], 0, 1) + f(dist2[0][prev], 0, 1) - 1,
  19. dp[2][prev] + f(dist1[0][prev], 1, 0) + f(dist2[0][prev], 1, 0) - 1
  20. );
  21.  
  22. dp[2][next] = max(
  23. dp[0][prev] + f(dist1[0][prev], 0, 0) + f(dist2[0][prev], 0, 0) - 1,
  24. dp[1][prev] + f(dist1[0][prev], 1, 0) + f(dist2[0][prev], 1, 0) - 1,
  25. dp[2][prev] + f(dist1[0][prev], 0, 1) + f(dist2[0][prev], 0, 1) - 1
  26. );
  27.  
  28. prev = next;
  29. }
  30.  
  31.  
  32. // чтоб посчитать, сколько вершин можно зохавать
  33. int f(int x, int d0, int d1) {
  34. return (x % 2 == 0) ? x-d0 : x-d1;
  35. }
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