제출 #995392

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
995392		asdfgrace	Stranded Far From Home (BOI22_island)	C++17	20 / 100	206 ms	33432 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;
    }
  }
  bool dif1 = true;
  vector<vector<int>> edges(n);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--; y--;
    pv(x); pv(y); pv(abs(x - y));
    if (abs(x - y) > 1) dif1 = false;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  pv(dif1);
  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';
  } else if (dif1) {
    /*
    sum(l + 1, r - 1) < both a[l] and a[r]
    so for each l find the min r so that sum(l + 1, r - 1) >= a[l]
    for each r find the max l so that sum(l + 1, r - 1) >= a[r]
    if mxl[r] <= l && mnr[l] >= r, they work probably
    so probably find the l with the max mnr
    or just an increasing seq of them that you can binary search
    so you have n ranges of invalid elems
    probably use set to remove all the elems
    
    implies that if we start in [6, 6] we can't escape to the left OR right
    */
    vector<long long> ps = s;
    for (int i = 1; i < n; i++) {
      ps[i] += ps[i - 1];
    }
    vector<int> mxl(n), mnr(n);
    for (int i = 1; i < n; i++) {
      if (i > 0) mxl[i] = upper_bound(ps.begin(), ps.end(), ps[i - 1] - s[i]) - ps.begin();
      if (i < n - 1) mnr[i] = lower_bound(ps.begin(), ps.end(), ps[i + 1] + s[i]) - ps.begin();
    }
    vector<int> ord(n);
    iota(ord.begin(), ord.end(), 0);
    sort(ord.begin(), ord.end(), [&] (int x, int y) {
      return mxl[x] > mxl[y];
    });
    int ptr = n - 1;
    vector<array<int, 2>> arr, bad;
    array<int, 2> tmp;
    for (int i = 0; i < n; i++) {
      pv(i); pv(ord[i]);
      while (ptr >= mxl[ord[i]]) {
        while (arr.size() && -arr.back()[0] <= mnr[ptr]) arr.pop_back();
        arr.push_back({-mnr[ptr], ptr});
        ptr--;
      }
      parr2d(arr);
      // GOAL: find the first >= i
      // ACTUAL GOAL: find the last <= -i
      tmp = {-mxl[ord[i]], (int) 1e9};
      int lb = upper_bound(arr.begin(), arr.end(), tmp) - arr.begin() - 1;
      if (lb != -1 && arr[lb][1] < ord[i] - 1) {
        bad.push_back({arr[lb][1] + 1, ord[i ] - 1});
        parr(bad.back());
      }
    }
    for (int i = n - 1; i > 0; i--) {
      if (s[i] > ps[i - 1]) {
        bad.push_back({0, i - 1});
        break;
      }
    }
    for (int i = 0; i < n - 1; i++) {
      if (s[i] > ps[n - 1] - ps[i]) {
        bad.push_back({i + 1, n - 1});
        break;
      }
    }
    parr2d(bad);
    set<int> st;
    for (int i = 0; i < n; i++) {
      st.insert(i);
    }
    for (auto [l, r] : bad) {
      auto it = st.lower_bound(l);
      while (it != st.end() && *it <= r) {
        st.erase(it);
        it = st.lower_bound(l);
      }
    }
    parr(mxl); parr(mnr); parr(ord);
    for (int i = 0; i < n; i++) {
      cout << (st.count(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