2019西北工业大学程序设计创新实践基地春季选拔赛 I Chino with Rewrite (并查集+树链剖分+线段树）

链接：https://ac.nowcoder.com/acm/contest/553/I

思路：离线整棵树，用并查集维护下联通的情况，因为值只有60个，用2的x(1<=x<=60)次方表示，树链剖分线段树区间取或维护，取得的值只要数二进制里面有多少个1就代表有多少个相同的数。

实现代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid ll m = (l + r) >> 1
const int M = 3e5+10;
ll sum[M<<2],f[M],head[M],son[M],siz[M],fa[M],dep[M];
ll cnt,cnt1,n,tid[M],top[M],a[M],op[M],u[M],v[M],ans[M];
struct node{
    ll to,next;
}e[M];
ll Find(ll x){
   if(x == f[x]) return x;
   return f[x] = Find(f[x]);
}

void add(ll u,ll v){
    e[++cnt1].to=v;e[cnt1].next=head[u];head[u]=cnt1;
    e[++cnt1].to=u;e[cnt1].next=head[v];head[v]=cnt1;
}
void dfs1(ll u,ll faz,ll deep){
    dep[u] = deep;
    siz[u] = 1;
    fa[u] = faz;
    //cout<<u<<" "<<faz<<" "<<deep<<endl;
    for(ll i = head[u];i ;i=e[i].next){
        ll v = e[i].to;
        if(v != fa[u]){
            dfs1(v,u,deep+1);
            siz[u] += siz[v];
            if(son[u]==-1||siz[v]>siz[son[u]])
                son[u] = v;
        }
    }
}

void dfs2(ll u,ll t){
    top[u] = t;
    tid[u] = cnt;
    //cout<<1<<endl;
    cnt++;
    if(son[u] == -1) return ;
    dfs2(son[u],t);
    for(ll i = head[u];i;i=e[i].next){
        ll v = e[i].to;
        if(v != son[u]&&v != fa[u])
            dfs2(v,v);
    }
}

void update(ll p,ll c,ll l,ll r,ll rt){
    if(l == r){
        sum[rt] = c;
        return ;
    }
    mid;
    if(p <= m) update(p,c,lson);
    else update(p,c,rson);
    sum[rt] = sum[rt<<1] | sum[rt<<1|1];
}

ll query(ll L,ll R,ll l,ll r,ll rt){
    if(L <= l&&R >= r){
         return sum[rt];
    }
    mid;
    ll ret = 0;
    if(L <= m) ret |= query(L,R,lson);
    if(R > m) ret |= query(L,R,rson);
    return ret;
}

ll solve(ll x){
    ll ans = 0;
    while(x){
        ans++;
        x -= (x&-x);
    }
    return ans;
}

ll ask(ll x,ll y){
    ll ans = 0;
    ll fx = top[x],fy = top[y];
    while(fx != fy){
        if(dep[fx] < dep[fy]) swap(x,y),swap(fx,fy);
        ans |= query(tid[fx],tid[x],1,n,1);
        x = fa[fx]; fx = top[x];
    }
    if(dep[x] > dep[y]) swap(x,y);
    ans |= query(tid[x],tid[y],1,n,1);
    return ans;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    ll q; cnt = 1;
    cin>>n>>q;
    memset(son,-1,sizeof(son));
    for(ll i = 1;i <= n;i ++){
        cin>>a[i];f[i] = i;
    }
    for(ll i = 1;i <= q;i ++){
        cin>>op[i]>>u[i]>>v[i];
        if(op[i] == 1){
            ll fx = Find(u[i]),fy = Find(v[i]);
            if(fx != fy) f[fx] = fy,add(u[i],v[i]),ans[i]=1;
        }
        else{
            ll fx = Find(u[i]),fy = Find(v[i]);
            if(fx != fy) ans[i] = -1;
        }
    }
    dfs1(1,0,1); dfs2(1,0);
    for(ll i = 1;i <= q;i ++){
        if(op[i] == 1&&ans[i]==1){
            ll num = (a[u[i]] + a[v[i]])/2;
            a[u[i]] = a[v[i]] = num;
            update(tid[u[i]],1LL<<num,1,n,1);
            update(tid[v[i]],1LL<<num,1,n,1);

        }
        else if(op[i] == 2&&ans[i]!=-1){
            ans[i] = solve(ask(u[i],v[i]));
        }
    }
    for(ll i = 1;i <= q;i ++)
        if(op[i] == 2) cout<<ans[i]<<endl;

}