标签:bzoj apio2012 启发式合并 平衡树 treap
题目大意:给出一棵树,每一个节点有两个值,分别是这个忍者的薪水和忍者的领导力。客户的满意程度是这个点的领导力乘能够取得人数,前提是取的人的薪水总和不超过总的钱数。
思路:只能在子树中操作,贪心的想,我们只要这个子树中cost最小的那些点就可以了。所以就深搜一次,每到一个节点上,把自己和所有子节点的平衡树启发式和并,然后保留不超过总钱数的人数,统计。数据范围比较大,能开long long的地方不要吝啬。
PS:吐槽一下,一开始这个题一直TTT,我以为是我常数写的太大了,别人都用左偏堆写,是不是平衡树已经成为了时代的眼泪了。。。后来我搞到了测点,跑了一下第一组数据等了1分多钟都没出解。我感觉我又要重写了,就出去转转。十分钟之后我回来发现居然出解了!而且居然还对了!!然后我仔细看了一遍程序。。。发现是启发式合并写反了。。。。写反了。。。反了。。。了。。。
CODE:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 100010 using namespace std; struct Complex{ long long cost,leader; }point[MAX]; struct Treap{ int random,size,cnt; long long val,sum; Treap *son[2]; Treap(long long _) { val = sum = _; size = cnt = 1; random = rand(); son[0] = son[1] = NULL; } int Compare(long long x) { if(x == val) return -1; return x > val; } void Maintain() { size = cnt; sum = val * cnt; if(son[0] != NULL) size += son[0]->size,sum += son[0]->sum; if(son[1] != NULL) size += son[1]->size,sum += son[1]->sum; } }*tree[MAX]; long long points,money; int head[MAX],total; int next[MAX << 1],aim[MAX << 1]; long long ans; inline void Add(int x,int y) { next[++total] = head[x]; aim[total] = y; head[x] = total; } inline void Rotate(Treap *&a,bool dir) { Treap *k = a->son[!dir]; a->son[!dir] = k->son[dir]; k->son[dir] = a; a->Maintain(),k->Maintain(); a = k; } inline void Insert(Treap *&a,long long x) { if(a == NULL) { a = new Treap(x); return ; } int dir = a->Compare(x); if(dir == -1) ++a->cnt; else { Insert(a->son[dir],x); if(a->son[dir]->random > a->random) Rotate(a,!dir); } a->Maintain(); } inline int FindMax(Treap *a) { return a->son[1] == NULL ? a->val:FindMax(a->son[1]); } inline void Delete(Treap *&a,long long x) { int dir = a->Compare(x); if(dir != -1) Delete(a->son[dir],x); else { if(a->cnt > 1) --a->cnt; else { if(a->son[0] == NULL) a = a->son[1]; else if(a->son[1] == NULL) a = a->son[0]; else { bool _ = (a->son[0]->random > a->son[1]->random); Rotate(a,_); Delete(a->son[_],x); } } } if(a != NULL) a->Maintain(); } void Transfrom(Treap *&from,Treap *&aim) { if(from == NULL) return ; Transfrom(from->son[0],aim); Transfrom(from->son[1],aim); for(int i = 1; i <= from->cnt; ++i) Insert(aim,from->val); delete from; from = NULL; } void DFS(int x) { tree[x] = new Treap(point[x].cost); if(point[x].cost <= money) ans = max(ans,(long long)point[x].leader); if(!head[x]) return ; for(int i = head[x]; i; i = next[i]) { DFS(aim[i]); if(tree[x]->size < tree[aim[i]]->size) swap(tree[x],tree[aim[i]]); Transfrom(tree[aim[i]],tree[x]); } while(tree[x]->sum > money) Delete(tree[x],FindMax(tree[x])); ans = max(ans,(long long)tree[x]->size * point[x].leader); } int main() { cin >> points >> money; for(int x,i = 1; i <= points; ++i) { scanf("%d%lld%lld",&x,&point[i].cost,&point[i].leader); Add(x,i); } DFS(0); cout << ans << endl; return 0; }
BZOJ 2809 APIO 2012 dispatching 平衡树启发式合并
标签:bzoj apio2012 启发式合并 平衡树 treap
原文地址:http://blog.csdn.net/jiangyuze831/article/details/40423821