USACO 2015 US Open, Silver Problem 3. Bessie's Birthday Buffet

原题下载

USACO2015OPEN-S3

答案

(Analysis by Nick Wu)

Start by sorting the patches of the grass in increasing order of quality.

Let f(i) be the maximum energy that we can accumulate if we end at patch i.

From patch i, we can compute the minimum distance from patch ii to every other patch. Then, for every patch j where patch j has lower quality grass than patch i, we have that f(i)≥f(j)+qj−E∗d(i,j).

It takes linear time to compute this information for a given patch. If we sort all the patches initially in O(Nlog⁡N), then this process takes O(N2), which will run in time.

Here is Mark Gordon's code.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

#define MAXN 1010

int Q[MAXN];
int DP[MAXN];
int D[MAXN];
vector<int> E[MAXN];

int main() {
  freopen("buffet.in", "r", stdin);
  freopen("buffet.out", "w", stdout);

  int N, ECST;
  cin >> N >> ECST;
  for (int i = 0; i < N; i++) {
    int D;
    cin >> Q[i] >> D;
    for (int j = 0; j < D; j++) {
      int v;
      cin >> v;
      E[i].push_back(v - 1);
    }
  }

  vector<int> PI;
  for (int i = 0; i < N; i++) {
    PI.push_back(i);
  }
  sort(PI.begin(), PI.end(), [&](int x, int y) {
    return Q[x] < Q[y];
  });

  int result = 0;
  for (int i = N - 1; i >= 0; i--) {
    int u = PI[i];

    queue<int> q;
    memset(D, -1, sizeof(D));
    q.push(u);
    D[u] = 0;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (int i = 0; i < E[v].size(); i++) {
        int nv = E[v][i];
        if (D[nv] == -1) {
          D[nv] = D[v] + 1;
          q.push(nv);
        }
      }
    }

    int res = Q[u];
    for (int j = 0; j < N; j++) {
      if (D[j] != -1) {
        res = max(res, Q[u] + DP[j] - ECST * D[j]);
      }
    }
    DP[u] = res;
    result = max(result, res);
  }

  cout << result << endl;
  return 0;
}

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