[Poi2011]Tree Rotations线段树合并

整理一下线段树合并的思路，大体是给每个树上节点分配一个根编号建一棵log长的权值线段树，一开始树上只有这个树节点的节点权

merge两个树节点的时候，对于当前合并的值域（例如两棵线段树的表示1到n/2的节点），

任意取两棵树中的一个节点编号，空的返回另一个，把树丰满起来，同时更新一下计数就可以了

#include<bits/stdc++.h>  
//#pragma comment(linker, "/STACK:1024000000,1024000000")   
#include<stdio.h>  
#include<algorithm>  
#include<queue>  
#include<string.h>  
#include<iostream>  
#include<math.h>                    
#include<stack>
#include<set>  
#include<map>  
#include<vector>  
#include<iomanip> 
#include<bitset>
using namespace std;         //#define ll long long  
#define ull unsigned long long
#define pb push_back  
#define FOR(a) for(int i=1;i<=a;i++) 
#define sqr(a) (a)*(a)
#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
ll qp(ll a,ll b,ll mod){ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
}
struct DOT{ll x;ll y;};
inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
const int dx[4]={0,0,-1,1};
const int dy[4]={1,-1,0,0};
const int inf=0x3f3f3f3f; 
const ll Linf=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;;const int maxn=8e6+34;int RT;
int n,a[maxn];int seg;
int tree[maxn],lson[maxn],rson[maxn];
int root[maxn];vector<int>G[maxn];
int Time;void pushup(int rt){tree[rt]=tree[lson[rt]]+tree[rson[rt]];}
void build(int &rt,int l,int r,int pos){rt=++seg;if(l==r){tree[rt]=1;return;}int m=l+r>>1;if(pos<=m)build(lson[rt],l,m,pos);else build(rson[rt],m+1,r,pos);pushup(rt);
}
ll ans,ans1,ans2;int merge(int x,int y){if(!x)return y;if(!y)return x;ans1+=1ll*tree[rson[x]]*tree[lson[y]];ans2+=1ll*tree[lson[x]]*tree[rson[y]];lson[x]=merge(lson[x],lson[y]);rson[x]=merge(rson[x],rson[y]);pushup(x);return x;
}void dfs(int u){if(a[u])return;dfs(G[u][0]);dfs(G[u][1]);ans1=ans2=0;root[u]=merge(root[G[u][0]],root[G[u][1]]);ans+=min(ans1,ans2);
}void init(int &rt){rt=++Time;scanf("%d",&a[Time]);if(a[Time])return;G[rt].pb(0);G[rt].pb(0);init(G[rt][0]);init(G[rt][1]);
}int main(){scanf("%d",&n);init(RT);for(int i=1;i<=Time;i++){if(a[i])build(root[i],1,n,a[i]);}dfs(RT);printf("%lld\n",ans);
}