BZOJ2809 [Apio2012]dispatching

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

 

BZOJ2809 [Apio2012]dispatching
  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\n", ans);
107     return 0;
108 }
View Code

 那个什么的扯淡教育部给我去死!!!

BZOJ2809 [Apio2012]dispatching

上一篇:转 Difference between WCF and Web API and WCF REST and Web Service


下一篇:WebMisSharp,WebMisCentral,企业框架正则表达式规则共享专版