给一棵N个点的树，然后问以某个点作为节点的时候，其子树中在购买力为M的情况下最多购买多少个节点，对于所有的子树，求节点数乘以该子树根节点的b值的最大值。
于是，想到的做法就是对于每个点去维护一个大根堆，如果大根堆内的和超过了M，那么就需要弹出最大的节点，于是，就是维护一个大根堆的合并了，合并的话，用左偏树最快了，于是我们在左偏树上pushup就可以了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <string> 5 #include <cstring> 6 #include <algorithm> 7 #include <limits> 8 #include <vector> 9 #include <stack> 10 #include <queue> 11 #include <set> 12 #include <map> 13 #include <bitset> 14 #include <unordered_map> 15 #include <unordered_set> 16 #define lowbit(x) ( x&(-x) ) 17 #define pi 3.141592653589793 18 #define e 2.718281828459045 19 #define INF 0x3f3f3f3f 20 #define HalF (l + r)>>1 21 #define lsn rt<<1 22 #define rsn rt<<1|1 23 #define Lson lsn, l, mid 24 #define Rson rsn, mid+1, r 25 #define QL Lson, ql, qr 26 #define QR Rson, ql, qr 27 #define myself rt, l, r 28 #define pii pair<int, int> 29 #define MP(a, b) make_pair(a, b) 30 using namespace std; 31 typedef unsigned long long ull; 32 typedef unsigned int uit; 33 typedef long long ll; 34 const int maxN = 1e5 + 7; 35 int N, root[maxN]; 36 ll M, b[maxN]; 37 struct node 38 { 39 ll val, sum; int d, c[2], fa, siz; 40 node(ll a=0, ll su=0, int b=0, int c1=0, int c2=0, int f=0, int s=0):val(a), sum(su), d(b), c{c1, c2}, fa(f), siz(s) {} 41 } t[maxN]; 42 int fid(int x) { return x == t[x].fa ? x : t[x].fa = fid(t[x].fa); } 43 void pushup(int x) 44 { 45 t[x].sum = t[t[x].c[0]].sum + t[t[x].c[1]].sum + t[x].val; 46 t[x].siz = t[t[x].c[0]].siz + t[t[x].c[1]].siz + 1; 47 } 48 int Merge(int x, int y) 49 { 50 if(!x || !y) return x | y; 51 if(t[x].val < t[y].val) swap(x, y); 52 t[x].c[1] = Merge(t[x].c[1], y); 53 t[t[x].c[1]].fa = x; 54 if(t[t[x].c[0]].d < t[t[x].c[1]].d) swap(t[x].c[0], t[x].c[1]); 55 t[x].d = t[t[x].c[1]].d + 1; 56 pushup(x); 57 return x; 58 } 59 int Pop(int x) 60 { 61 int f = fid(x); 62 while(t[f].sum > M) 63 { 64 t[t[f].c[0]].fa = t[f].c[0]; 65 t[t[f].c[1]].fa = t[f].c[1]; 66 t[f].fa = Merge(t[f].c[0], t[f].c[1]); 67 f = t[f].fa; 68 } 69 return f; 70 } 71 vector<int> to[maxN]; 72 ll ans = 0; 73 void dfs(int u) 74 { 75 int fu, fv; root[u] = u; 76 for(int v : to[u]) 77 { 78 dfs(v); 79 fu = fid(root[u]); fv = fid(root[v]); 80 root[u] = Merge(fu, fv); 81 } 82 root[u] = Pop(fid(u)); 83 ans = max(ans, t[root[u]].siz * b[u]); 84 } 85 int main() 86 { 87 scanf("%d%lld", &N, &M); 88 for(int i=1, ff; i<=N; i++) 89 { 90 scanf("%d%lld%lld", &ff, &t[i].val, &b[i]); 91 t[i].fa = i; t[i].siz = 1; t[i].sum = t[i].val; 92 if(ff) to[ff].push_back(i); 93 } 94 dfs(1); 95 printf("%lld\n", ans); 96 return 0; 97 }
