November 2013 Bronze Problem 2: Goldilocks and the N Cows

Result: 7/10 *******ttt

I think that testing every boundary line might be slow, but it's the only way I can think of that does it completely.

milktemp.cpp

#include <iostream>
#include <fstream>

#include <set>

#define loopi(x) loop_start(i,x,0)
#define loopj(x) loop_start(j,x,0)
#define loop_start(i, x,s) for(int i = (s); i < (x); ++i)
#define loop_rev(a,x,stop) for(int a = (x); a >= (stop); --a)

using namespace std;

int N, X, Y, Z;

int main() {
 ofstream fout ("milktemp.out");
 ifstream fin ("milktemp.in");
 // get input
 fin >> N >> X >> Y >> Z;
 // calculate
 int *a = new int[N], *b = new int[N];
 set<int> boundaries;
 loopi(N) {
  fin >> a[i] >> b[i];
  boundaries.insert(a[i]);
  boundaries.insert(b[i]);
 }
 int best_score = 0, best_suitable = 0;
 for (set<int>::iterator it = boundaries.begin(); it != boundaries.end(); ++it) {
  int too_low = 0, too_high = 0, suitable = 0;
  loopi(N) {
   if(a[i] > *it)
    ++too_low;
   else if(b[i] < *it)
    ++too_high;
   else
    ++suitable;
  }
  // The best solution has as many as possible in the suitable range
  if(suitable < best_suitable)
   continue;
  best_suitable = suitable;
  // Update best score
  const int new_score = X * too_low + Y * suitable + Z * too_high;
  if(new_score > best_score)
   best_score = new_score;
 }
 // Write output
 fout << best_score;
 // Clean up
 delete[] a;
 delete[] b;
 return 0;
}