BZOJ 2333 SCOI 2011 棘手的操作 可并堆

做此题的原因

• 题号美

题目大意

给出一个序列，支持一堆操作（具体看下面）。让你维护它。

思路

U x y：我们需要可并堆来将两个堆合并。
A1 x v：将这个点从堆中拽出来，改了之后再合并回去。
A2 x v：在堆顶打标记。
A3：记录一个全局变量记录。
F1 x：将这个点到堆顶的链上的所有标记下传，之后返回自己的大小。
F2 x：返回堆顶。
F3：用一个堆（set也行）维护所有堆顶的元素。需要仔细讨论一下。

CODE

#define _CRT_SECURE_NO_WARNINGS

#include <set>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 300010
using namespace std;

multiset<int> G;

struct Heap{
    int val,plus;
    Heap *son[2],*father;

    Heap(int _):val(_) {
        son[0] = son[1] = father = NULL;
        plus = 0;
    }
    Heap() {}
    void Plus(int _) {
        val += _;
        plus += _;
    }
    void PushDown() {
        if(plus) {
            if(son[0] != NULL)  son[0]->Plus(plus);
            if(son[1] != NULL)  son[1]->Plus(plus);
            plus = 0;
        }
    }
}heap[MAX];

Heap *Find(Heap *x)
{
    while(x->father != NULL)
        x = x->father;
    return x;
}

inline Heap *Merge(Heap *x,Heap *y)
{
    if(x == NULL)   return y;
    if(y == NULL)   return x;
    if(x->val < y->val) swap(x,y);
    x->PushDown();
    bool k = rand()&1;
    x->son[k] = Merge(x->son[k],y);
    x->son[k]->father = x;
    return x;
}

inline void Union(Heap *x,Heap *y)
{
    G.erase(G.find(min(x->val,y->val)));
    Merge(x,y);
}

inline void ClearTag(Heap *x)
{
    static Heap *stack[MAX];
    int top = 0;
    while(x != NULL)
        stack[++top] = x,x = x->father;
    for(int i = top; i; --i)
        stack[i]->PushDown();
}

inline void PlusPoint(Heap *x,int c)
{   
    if(x->father == NULL) {
        G.erase(G.find(x->val));
        x->PushDown();
        Heap *temp = Merge(x->son[0],x->son[1]);
        if(temp != NULL)
            temp->father = NULL;
        x->val += c;
        x->son[0] = x->son[1] = NULL;
        G.insert(Merge(x,temp)->val);
        return ;
    }
    Heap *fx = Find(x);
    G.erase(G.find(fx->val));
    ClearTag(x);
    Heap *temp = Merge(x->son[0],x->son[1]);
    bool k = x->father->son[1] == x;
    x->father->son[k] = temp;
    if(temp != NULL)
        temp->father = x->father;
    x->father = NULL;
    x->val += c;
    x->son[0] = x->son[1] = NULL;
    G.insert(Merge(fx,x)->val);
}

inline void PlusBlock(Heap *x,int c)
{
    Heap *fx = Find(x);
    G.erase(G.find(fx->val));
    fx->Plus(c);
    G.insert(fx->val);
}

int g_add;

inline void PlusAll(int c)
{
    g_add += c;
}

inline int AskPoint(Heap *x)
{
    ClearTag(x);
    return x->val + g_add;
}

inline int AskBlock(Heap *x)
{
    Heap *fx = Find(x);
    return fx->val + g_add;
}

inline int AskAll()
{
    multiset<int>::iterator it = G.end();
    return *(--it) + g_add;
}

int points,asks;
char s[10];

int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    cin >> points;
    for(int x,i = 1; i <= points; ++i) {
        scanf("%d",&x);
        heap[i] = Heap(x);
        G.insert(x);
    }
    cin >> asks;
    for(int x,y,i = 1; i <= asks; ++i) {
        scanf("%s",s);
        if(s[0] == ‘U‘) {
            scanf("%d%d",&x,&y);
            Heap *fx = Find(&heap[x]),*fy = Find(&heap[y]);
            if(fx == fy)    continue;
            Union(fx,fy);
        }
        if(s[0] == ‘A‘) {
            if(s[1] == ‘1‘) {
                scanf("%d%d",&x,&y);
                PlusPoint(&heap[x],y);
            }
            if(s[1] == ‘2‘) {
                scanf("%d%d",&x,&y);
                PlusBlock(&heap[x],y);
            }
            if(s[1] == ‘3‘) {
                scanf("%d",&x);
                PlusAll(x);
            }
        }
        if(s[0] == ‘F‘) {
            if(s[1] == ‘1‘) {
                scanf("%d",&x);
                printf("%d\n",AskPoint(&heap[x]));
            }
            if(s[1] == ‘2‘) {
                scanf("%d",&x);
                printf("%d\n",AskBlock(&heap[x]));
            }
            if(s[1] == ‘3‘)
                printf("%d\n",AskAll());
        }
    }
    return 0;
}