To solve this problem, you construct best solutions for "prefixes" of the fence, and extend them by individual painters one by one. You can consider the painters by end panel left-to-right (ordering painters with the same end panel arbitrarily).

We will store the maximum number of panels painted for our prefixes for simplicity, and subtract this from the total number of panels when printing the result.

Then you iterate through the painters, and you calculate for each painter what the best value would be for that painter's end panel, both if the painter painted and if the painter did not paint.

best = 0
for (p1 : painter)
   best max= p1.right - p1.left + 1
   for (p2 : painters left of p1)
      if (p2.right < p1.left)
         best max= p2.best + p1
   p1.best = best

This is quadratic, which is not quite fast enough. You can improve the time easily by, instead of iterating through all painters, find the maximum value strictly less than the left side of the current painter using a HashMap or C++ map. This is the same solution as the previous one, but we only consider the prefix that is closest to this painter's left boundary.

best = 0
dmap[0] = 0
for (p1 : painter)
   // find the best solution so far to the *left* of this guy's left boundary
   int bestprev = dmap.upper_bound(p1.left) ;
   // check if that plus this guy's span is better than what we've seen
   best max= p1.right - p1.left + 1 + dmap[bestprev]
   // add it to the map.
   dmap[p1.lright] = best ;