看来蒟蒻我还是直接退役算了。。。
此题就是维护子树的和，删除子树中当前最大元素，并且可以合并两个子树信息，想到了左偏树。。。
做完了233
1 /************************************************************** 2 Problem: 2809 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:804 ms 7 Memory:7624 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 const int N = 100005; 16 17 struct heap{ 18 int v, l, r, dep; 19 }h[N]; 20 int cnt_heap; 21 22 struct edge { 23 int next, to; 24 edge() {} 25 edge(int _n, int _t) : next(_n), to(_t) {} 26 } e[N]; 27 int first[N], tot; 28 29 struct tree_node { 30 int cost, l, root; 31 ll sum, sz; 32 } tr[N]; 33 34 int n, m; 35 ll ans; 36 37 inline int read() { 38 int x = 0, sgn = 1; 39 char ch = getchar(); 40 while (ch < '0' || '9' < ch) { 41 if (ch == '-') sgn = -1; 42 ch = getchar(); 43 } 44 while ('0' <= ch && ch <= '9') { 45 x = x * 10 + ch - '0'; 46 ch = getchar(); 47 } 48 return sgn * x; 49 } 50 51 inline void add_edge(int x, int y) { 52 e[++tot] = edge(first[x], y); 53 first[x] = tot; 54 } 55 56 inline int new_heap(int x) { 57 h[++cnt_heap].v = x; 58 h[cnt_heap].l = h[cnt_heap].r = h[cnt_heap].dep = 0; 59 return cnt_heap; 60 } 61 62 int Merge(int x, int y) { 63 if (!x || !y) return x + y; 64 if (h[x].v < h[y].v) 65 swap(x, y); 66 h[x].r = Merge(h[x].r, y); 67 if (h[h[x].l].dep < h[h[x].r].dep) 68 swap(h[x].l, h[x].r); 69 h[x].dep = h[h[x].r].dep + 1; 70 return x; 71 } 72 73 inline int Top(int p) { 74 return h[p].v; 75 } 76 77 void Pop(int &p) { 78 p = Merge(h[p].l, h[p].r); 79 } 80 81 void dfs(int p) { 82 int x, y; 83 tr[p].root = new_heap(tr[p].cost); 84 tr[p].sum = tr[p].cost, tr[p].sz = 1; 85 for (x = first[p]; x; x = e[x].next) { 86 dfs(y = e[x].to); 87 tr[p].sum += tr[y].sum, tr[p].sz += tr[y].sz; 88 tr[p].root = Merge(tr[p].root, tr[y].root); 89 } 90 while (tr[p].sum > m) { 91 tr[p].sum -= Top(tr[p].root), Pop(tr[p].root); 92 --tr[p].sz; 93 } 94 ans = max(ans, tr[p].sz * tr[p].l); 95 } 96 97 int main() { 98 int i, x; 99 n = read(), m = read(); 100 for (i = 1; i <= n; ++i) { 101 x = read(); 102 add_edge(x, i); 103 tr[i].cost = read(), tr[i].l = read(); 104 } 105 dfs(1); 106 printf("%lld ", ans); 107 return 0; 108 }
那个什么的扯淡教育部给我去死!!!