USACO 2016 February Contest, Gold Problem 3. Fenced In

Category: 計算機國際競賽 Date: 2017年11月9日上午10:37

原題下載

USACO2016-FEB-G3

答案

(Analysis by Nathan Pinsker)

For this problem, we're given a grid and want to connect all the interiors to each other. If we think of each interior region as a point and add edges between points representing the length of the fence that separates them, then this problem becomes one of finding a minimum spanning tree on a graph. Since our graph has? $O (n * m)$ vertices and each vertex has at most 4 edges, it also has? $O (n * m)$ ?edges, meaning that we can simply run our favorite minimum spanning tree algorithm to solve the problem.

However, we can actually make this solution even faster! If you're interested, see the platinum version of this problem.

Here is Mark Gordon's code, with an implementation of Kruskal's algorithm:

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int A, B, N, M;

int nd(int r, int c) {
  return r * M + c;
}

int P[2010*2010];

int find(int x) {
  return x == P[x] ? x : P[x] = find(P[x]);
}

bool merge(int x, int y) {
  int X = find(x);
  int Y = find(y);
  if (X == Y) return false;
  P[X] = P[Y] = P[x] = P[y] = rand() % 2 ? X : Y;
  return true;
}

int main() {
  cin >> A >> B >> N >> M;

  vector<int> VA(N + 1), HA(M + 1);
  for (int i = 0; i < N; i++) {
    cin >> VA[i];
  }
  for (int j = 0; j < M; j++) {
    cin >> HA[j];
  }

  sort(VA.begin(), VA.end());
  for (int i = 0; i < N; i++) {
    VA[i] = VA[i + 1] - VA[i];
  }
  VA[N] = B - VA[N];

  sort(HA.begin(), HA.end());
  for (int i = 0; i < M; i++) {
    HA[i] = HA[i + 1] - HA[i];
  }
  HA[M] = A - HA[M];

  sort(VA.begin(), VA.end());
  sort(HA.begin(), HA.end());
  N++; M++;

  for (int i = 0; i < N * M; i++) {
    P[i] = i;
  }

  long long result = 0;
  for (int vi = 0, hi = 0; vi < N || hi < M; ) {
    if (hi == M || (vi < N && VA[vi] < HA[hi])) {
      for (int i = 0; i + 1 < M; i++) {
        if (merge(nd(vi, i), nd(vi, i + 1))) {
          result += VA[vi];
        }
      }
      vi++;
    } else {
      for (int i = 0; i + 1 < N; i++) {
        if (merge(nd(i, hi), nd(i + 1, hi))) {
          result += HA[hi];
        }
      }
      hi++;
    }
  }
  cout << result << endl;

  return 0;
}

以上就是關于【USACO 2016 February Contest, Gold Problem 3. Fenced In】的解答，如需了解學校/賽事/課程動態，可至翰林教育官網獲取更多信息。

外灘大會見證AI爆發！緊跟智能時代，從USACO計算機奧賽起步！