Garland

题目地址:

https://ac.nowcoder.com/acm/problem/111886

基本思路:

首先我们知道只要整棵树的温度总和是确定的,
那么这个连通块的温度就也是确定的,
考虑树形,令表示以为根的子树的温度总和,
只要子树温度和等于,
那么我们就将这棵子树断开,消去这部分贡献,然后记录断点。
只要符合条件的断点数目大于等于，那么就是符合的。

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define int long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1e6 + 10;
struct Edge{
    int to,next;
}edge[maxn << 1];
int cnt = 0,head[maxn],w[maxn];
void add_edge(int u,int v){
  edge[++cnt].next = head[u];
  edge[cnt].to = v;
  head[u] = cnt;
}
int f[maxn],sum,tmp,tot,rt,ans[maxn];
void dfs(int u,int par) {
  f[u] = w[u];
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    if (to == par) continue;
    dfs(to, u);
    f[u] += f[to];
  }
  if (f[u] == tmp && u != rt) ans[++tot] = u,f[u] = 0; // 特别注意断点不能为根;
}
int n,x,t;
signed main() {
  IO;
  cin >> n;
  cnt = 0,sum = 0; mset(head,-1);
  rep(i,1,n){
    cin >> x >> w[i];
    sum += w[i];
    if(x == 0) rt = i;
    else add_edge(x,i),add_edge(i,x);
  }
  if(sum % 3 != 0) {
    cout << -1 << '\n';
    return 0;
  }
  tmp = sum / 3,tot = 0;
  dfs(rt,0);
  if(tot >= 2){
    cout << ans[2] << ' ' << ans[1] << '\n';
  }else cout << -1 << '\n';
  return 0;
}