BZOJ 3673 可持久化并查集 by zky && 3674 可持久化并查集加强版

标签：bzoj   可持久化线段树   可持久化并查集   可持久化数组   

题目大意：维护一种数据结构实现可持久化并查集。

思路：利用可持久化线段树实现可持久化数组维护可持久化并查集。（不知道3674哪里加强了。。。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define RANGE 8000010
#define MAX 200200
using namespace std;
 
struct SegTree{
    SegTree *son[2];
    int val;
     
    void *operator new(size_t,SegTree *_,SegTree *__,int ___);
}mempool[RANGE],*C = mempool,*father[MAX],*h[MAX];
void *SegTree:: operator new(size_t,SegTree *_,SegTree *__,int ___) {
    C->son[0] = _;
    C->son[1] = __;
    C->val = ___;
    return C++;
}
 
int cnt,asks;
int verson[MAX],now_ver,latest;
int last_ans;
 
SegTree *Modify(SegTree *a,int l,int r,int x,int val)
{
    if(l == r)  return new(NULL,NULL,val)SegTree;
    int mid = (l + r) >> 1;
    if(x <= mid) return new(Modify(a->son[0],l,mid,x,val),a->son[1],0)SegTree;
    return new(a->son[0],Modify(a->son[1],mid + 1,r,x,val),0)SegTree;
}
 
int Ask(SegTree *a,int l,int r,int x)
{
    if(l == r)  return a->val;
    int mid = (l + r) >> 1;
    if(x <= mid) return Ask(a->son[0],l,mid,x);
    return Ask(a->son[1],mid + 1,r,x);
}
 
inline int Find(int x)
{
    int y;
    while(x)
        x = Ask(father[now_ver],1,cnt,y = x);
    return y;
}
 
inline void Unite(int x,int y)
{
    int fx = Find(x),fy = Find(y);
    if(fx == fy)    return ;
    ++latest;
    int hx = Ask(h[now_ver],1,cnt,fx);
    int hy = Ask(h[now_ver],1,cnt,fy);
    if(hx < hy)
        swap(x,y),swap(fx,fy),swap(hx,hy);
    father[latest] = Modify(father[now_ver],1,cnt,fy,fx);
    h[latest] = Modify(h[now_ver],1,cnt,fx,hx + hy);
    now_ver = latest;
}
 
inline bool Query(int x,int y)  
{  
    return Find(x) == Find(y);  
}  
 
int main()
{
    cin >> cnt >> asks;
    father[0] = new(C,C,0)SegTree;
    h[0] = new(C,C,1)SegTree;
    for(int flag,x,y,i = 1; i <= asks; ++i) {
        scanf("%d",&flag);
        if(flag == 1) {
            scanf("%d%d",&x,&y);
            x ^= last_ans,y ^= last_ans;
            Unite(x,y);
        }
        else if(flag == 2) {
            scanf("%d",&x);
            x ^= last_ans;
            now_ver = verson[x];
        }
        else {
            scanf("%d%d",&x,&y);
            x ^= last_ans,y ^= last_ans;
            printf("%d\n",last_ans = Query(x,y));
        }
        verson[i] = now_ver;
    }
    return 0;
}

BZOJ 3673 可持久化并查集 by zky && 3674 可持久化并查集加强版

标签：bzoj   可持久化线段树   可持久化并查集   可持久化数组   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41692063