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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#186183#6425. Harmonious RectangleshimemingRE 0ms0kbC++202.3kb2023-09-23 11:28:062023-09-23 11:28:06

Judging History

你现在查看的是最新测评结果

  • [2023-09-23 11:28:06]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2023-09-23 11:28:06]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define debug(x) {cerr << #x << " = " << x << '\n'; }

#define X first
#define Y second
#define pb push_back
#define int long long

const int MAXN = 1e5+5;
int n;
ll a[MAXN];
int t[MAXN];
vector<vector<int>> G;
vector<ll> dp;

ll dfs(int ind, int p) {
  assert(n!=1);
  if (ind != 0 && G[ind].size() == 1) {
    return 0;
  }
  if (dp[ind] != 0) return dp[ind];
  ll rt = -1;
  ll sum = 0, maxn = -1, maxn31 = -1, maxn32 = -1;
  vector<pair<int, ll>> sub;
  for (int i : G[ind]) {
    if (i == p) continue;
    int tmp = dfs(i, ind);
    sub.pb({i, tmp});
    sum += tmp;
    if (maxn == -1 || a[i] > a[maxn]) maxn = i;
    if (t[i] == 3) {
      if (maxn32 == -1 || a[i] >= a[maxn32]) {
        maxn32 = i;
        if (maxn31 == -1 || a[maxn32] >= a[maxn31]) swap(maxn31, maxn32);
      }
    }
  }
  rt = sum + a[maxn];
  // debug(ind);
  // debug(rt);
  if (G[ind].size() > 2) {
    // debug(ind);
    int s1 = -1, s2 = -1;
    auto diff = [&](int k) { return a[sub[k].X]-sub[k].Y; };
    for (int i = 0; i < (int)sub.size(); i++) {
      if (s2 == -1 || diff(i) >= diff(s2)) {
        s2 = i;
        if (s1 == -1 || diff(s2) >= diff(s1)) swap(s1, s2);
      }
    }
    if (maxn31 != -1) {
      // debug(maxn31);
      // debug(sub[s1].X);
      // debug(sub[s2].X);
      if (sub[s1].X != maxn31) {
        rt = max(rt, sum+a[maxn31]+a[sub[s1].X]-sub[s1].Y);
      } else {
        rt = max(rt, sum+a[maxn31]+a[sub[s2].X]-sub[s2].Y);
        if (maxn32 != -1) {
          rt = max(rt, sum+a[maxn32]+a[sub[s1].X]-sub[s1].Y);
        }
      }
    }
  }
  dp[ind] = rt;
  return rt;
}

signed main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T; cin >> T;
  while (T--) {
    cin >> n;
    G.resize(n+1);
    dp.resize(n+1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> t[i];
    for (int i = 1; i <= n-1; i++) {
      int x, y; cin >> x >> y;
      G[x].pb(y);
      G[y].pb(x);
    }
    if (n == 1) {
      cout << a[1] << '\n';
      continue;
    }
    t[1] = 2;
    G[0].pb(1);
    G[1].pb(0);
    dfs(0, -1);
    cout << dp[0] << '\n';
    G.clear();
    dp.clear();
  }
  return 0;
}

详细

Test #1:

score: 0
Runtime Error

input:

3
1 4
2 2
3 3

output:


result: