Submission #995374

#	Time	Username	Problem	Language	Result	Execution time	Memory
995374		asdfgrace	Stranded Far From Home (BOI22_island)	C++17	20 / 100	206 ms	37972 KiB

island

#include <bits/stdc++.h>
using namespace std;

#define dbg(x) x
#define prt(x) dbg(cerr << x)
#define pv(x) prt(#x << " = " << x << '\n')
#define parr(x) dbg(prt(#x << " = "); for (auto y : x) prt(y << ' '); prt('\n'));
#define parr2d(x) dbg(prt(#x << " = \n"); for (auto y : x) parr(y); prt('\n'));

/*
which ties are possible?
a can convince b if sum(a) > sum(b)
note that nodes are weighted
note that the tie with the strictly least number can't be 1 in the end
one with strictly most can be 1 in the end
how to greedily make 1 the largest?
dfs from the node
and keep adding the smallest adj greedily
and make sure that the size of this color's comp is greater than any other color's comp

subtask 1: brute force
subtask 2: tree; every node's parent has greater value than self
you can always get everything in your own subtree
go to the root
and for every node on the path to the root
you also absorb the parent
and everything else in the parent's subtree - note that that is guaranteed to be possible
if you've already absorbed the parent
this dp cond is checkable as you go up the tree
(this node's subtree sum >= parent's value?)
all the values to the root have to be true
for a node to work
subtask 3: just some weird stuff with set ig
*/

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<long long> s(n);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  bool inc = false;
  for (int i = 0; i < n - 1; i++) {
    if (s[i + 1] > s[i]) {
      inc = true;
    }
  }
  vector<vector<int>> edges(n);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--; y--;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  if (n <= 2000 && m <= 2000) {
    for (int i = 0; i < n; i++) {
      vector<bool> vis(n, false);
      vis[i] = true;
      priority_queue<array<long long, 2>> que;
      que.push({-s[i], i});
      bool ok = true;
      long long tot = s[i];
      while (que.size()) {
        long long sz = -que.top()[0], node = que.top()[1];
        que.pop();
        if (sz > tot) {
          ok = false;
          break;
        }
        if (node != i) {
          tot += sz;
        }
        for (auto next : edges[node]) {
          if (!vis[next]) {
            vis[next] = true;
            que.push({-s[next], next});
          }
        }
      }
      cout << (ok ? 1 : 0);
    }
    cout << '\n';
  } else if (m == n - 1 && !inc) {
    vector<bool> ok(n, true);
    vector<long long> sum = s;
    function<void(int, int)> dfs1 = [&] (int node, int par) {
      for (auto next : edges[node]) {
        if (next != par) {
          dfs1(next, node);
          sum[node] += sum[next];
        }
      }
    };
    dfs1(0, 0);
    function<void(int, int)> dfs2 = [&] (int node, int par) {
      if (node != 0) {
        ok[node] = ok[par];
        if (s[par] > sum[node]) {
          ok[node] = false;
        }
      }
      for (auto next : edges[node]) {
        if (next != par) {
          dfs2(next, node);
        }
      }
    };
    dfs2(0, 0);
    for (int i = 0; i < n; i++) {
      cout << (ok[i] ? 1 : 0);
    }
    cout << '\n';
  }
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 30
• Subtask 5: 0 / 35