可並堆 - Monkey King（HDU1512）

傳送門

Analysis

我都忘了自己學過左偏樹了【嗚嗚嗚嗚】

就當模板複習吧

合併操作

int merge(int x,int y){
	if(!x||!y) return x+y;
	if(tree[x].val<tree[y].val) swap(x,y);//這裏處理的是大根堆
	tree[x].rc=merge(tree[x].rc,y);
	fa[tree[x].rc]=x;//注意更新fa
	if(tree[tree[x].lc].dep<tree[tree[x].rc].dep) swap(tree[x].lc,tree[x].rc);
	if(tree[x].rc) tree[x].dep=tree[tree[x].rc].dep+1;//斜深度定義爲根到右兒子的步數
	else tree[x].dep=0;
	return x;
}

尋找堆頂

int find(int x){
	int rt=x;
	while(rt!=fa[rt]) rt=fa[rt];
	int f=fa[x];
	while(f!=rt){
		fa[x]=rt;
		x=f;
		f=fa[x];//有點像並查集的路徑壓縮
	}
	return rt;
}

彈出堆頂，並且將剩下的部分合並

int pop(int x){//先確定x爲當前堆頂
	int l=tree[x].lc;
	int r=tree[x].rc;
	fa[l]=l;fa[r]=r;
	tree[x].lc=tree[x].rc=tree[x].dep=0;
	return merge(l,r);
}

Code

完整代碼

#include<bits/stdc++.h>
#define re register
#define in read()
using namespace std;
inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?(exit(0),EOF):*p1++;
}
inline int read(){
	char ch;int f=1,res=0;
	while((ch=nc())<'0'||ch>'9') if(ch=='-') f=-1;
	while(ch>='0'&&ch<='9'){
		res=(res<<1)+(res<<3)+(ch^48);
		ch=nc();
	}
	return f==1?res:-res;
}
const int N=100009;
struct node{
	int lc,rc,dep,val;
}tree[N];
int fa[N];
int merge(int x,int y){
	if(!x||!y) return x+y;
	if(tree[x].val<tree[y].val) swap(x,y);
	tree[x].rc=merge(tree[x].rc,y);
	fa[tree[x].rc]=x;///
	if(tree[tree[x].lc].dep<tree[tree[x].rc].dep) swap(tree[x].lc,tree[x].rc);
	if(tree[x].rc) tree[x].dep=tree[tree[x].rc].dep+1;
	else tree[x].dep=0;
	return x;
}
int find(int x){
	int rt=x;
	while(rt!=fa[rt]) rt=fa[rt];
	int f=fa[x];
	while(f!=rt){
		fa[x]=rt;
		x=f;
		f=fa[x];
	}
	return rt;
}
int pop(int x){
	int l=tree[x].lc;
	int r=tree[x].rc;
	fa[l]=l;fa[r]=r;
	tree[x].lc=tree[x].rc=tree[x].dep=0;
	return merge(l,r);
}
int main(){
//	freopen("data.in","r",stdin);
	int n,m; 
	while((n=in)!=EOF){
		memset(tree,0,sizeof(tree));
		for(re int i=1;i<=n;++i) {
			tree[i].val=in;
			fa[i]=i;
		}
		m=in;
		for(re int i=1;i<=m;++i){
			int x=in,y=in;
			int fx=find(x),fy=find(y);
			if(fx==fy){printf("-1\n");continue;}
			int X=pop(fx);
			tree[fx].val/=2;
			X=merge(X,fx);
			int Y=pop(fy);
			tree[fy].val/=2;
			Y=merge(Y,fy);
			printf("%d\n",tree[merge(X,Y)].val);
		}
	}
	return 0;
}