USACO 2016 February Contest, Platinum Problem 1. Load Balancing

原题下载

USACO2016-FEB-P1

答案

(Analysis by Nathan Pinsker)

The most immediately obvious solution is to try all possible locations for one gate (let's arbitrarily assume the north-south gate), then try all possible locations for the other gate; unfortunately, this takes O(n2)O(n2) time in total and is too slow.

However, we can speed up figuring out where to put the east-west gate, given the position of the north-south gate already. As we move our east-west gate from top to bottom, note that the number of cows in each of the northern two regions is a nondecreasing function. Similarly, the number of cows in each of the southern two regions is a nonincreasing function. Thus, max(cows in northwest region, cows in northeast region) is also nondecreasing, and max(cows in southwest region, cows in southeast region) is also nonincreasing.

This means that we can use a form of binary search to figure out where they cross: we simply look for the y-value where the two graphs of these functions cross. Put more simply, if there are too many cows in the top regions, then we need to move our fence downwards, and if there are too many cows in the bottom regions, then we need to move our fence upwards. Note that as we iterate over all possible positions for the north-south gate, we need to efficiently keep track of cows as they move from one side of the fence to the other. We can do this using a segment tree, which also lets us compute the number of cows in each region very easily.

Here is Mark Gordon's code, which implements this idea:

#include <iostream>
#include <vector>
#include <queue>
#include <functional>
#include <set>
#include <algorithm>
#include <map>

using namespace std;

#define MAXN (1 << 17)

int S1[2 * MAXN];
int S2[2 * MAXN];

void update(int* S, int x, int v) {
  x += MAXN;
  S[x] += v;
  for (x = x >> 1; x; x = x >> 1) {
    S[x] = S[2 * x + 0] + S[2 * x + 1];
  }
}

int query(int x, int l1, int r1, int l2, int r2) {
  if (x > MAXN) {
    (l1 < r1 ? l1 : r1) += S1[x];
    (l2 < r2 ? l2 : r2) += S2[x];
    return max(max(l1, r1), max(l2, r2));
  }

  int c1 = x * 2 + 0;
  int c2 = x * 2 + 1;
  if (max(l1 + S1[c1], l2 + S2[c1]) < max(r1 + S1[c2], r2 + S2[c2])) {
    return query(c2, l1 + S1[c1], r1, l2 + S2[c1], r2);
  } else {
    return query(c1, l1, r1 + S1[c2], l2, r2 + S2[c2]);
  }
}

int main() {
  int N; cin >> N;

  vector<pair<int, int> > A(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i].first >> A[i].second;
  }
  sort(A.begin(), A.end());

  int xp = 0, lx = A[0].first;
  for (int i = 0; i < N; i++) {
    if (A[i].first != lx) {
      lx = A[i].first;
      xp++;
    }
    A[i].first = xp;
  }

  sort(A.begin(), A.end(), [](pair<int, int> a, pair<int, int> b) {
    return make_pair(a.second, a.first) < make_pair(b.second, b.first);
  });

  for (int i = 0; i < N; i++) {
    update(S1, A[i].first, 1);
  }

  int result = N;
  for (int i = 0; i < N; ) {
    int sz = 0;
    while (i + sz < N && A[i].second == A[i + sz].second) {
      update(S1, A[i + sz].first, -1);
      update(S2, A[i + sz].first, 1);
      sz++;
    }
    result = min(result, query(1, 0, 0, 0, 0));
    i += sz;
  }
  cout << result << endl;
}
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