Bzoj3786: 星系探索——Splay

题面

　　 Bzoj3786

解析

　　 上课讲稿上的例题

　　这道题是套路题，是括号序的应用，进入节点时打上$+1$标记, 退出时打上$-1$标记，这个是作为点权的系数
　　先看操作2， 需要更改父节点，就是把一段区间提取出来，插入另一个地方，显然可以用Splay维护，先提取区间，再把新父亲的$+1$点旋转至根，把区间挂在根的后继的左儿子上，再把这个节点旋转至根，以更新信息

　　对于操作1，求点到根的路径和，就是求括号序列的前缀和，该点对应的$+1$点或$-1$点的前缀和都可，我是把$-1$的点旋转至根，答案就是根的左儿子的子树和，因此需要维护子树和

　　最后还有操作3，显然打标记，问题在对于子树和的改变， 记该子树中$+1$标记有$cnt1$个，$-1$标记有$cnt_1$个，则子树和需要加$(cnt1 - cnt_1) * delta$，因此需要维护子树中$+1$与$-1$的个数

　代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 100004;

template<class T> void read(T &re)
{
    re=0;
    T sign=1;
    char tmp;
    while((tmp=getchar())&&(tmp<'0'||tmp>'9')) if(tmp=='-') sign=-1;
    re=tmp-'0';
    while((tmp=getchar())&&(tmp>='0'&&tmp<='9')) re=(re<<3)+(re<<1)+(tmp-'0');
    re*=sign;
}

int n, m, root, cnt, a[maxn], fir[maxn], sec[maxn];
vector<int> G[maxn];

struct spaly_tree{
    int s[2], fa, val, add, cnt1, cnt_1, fl[2];
    ll sum;
}tr[maxn<<1];

void dfs(int x)
{
    fir[x] = ++cnt;
    tr[cnt].fl[1] = 1;
    tr[cnt].val = a[x];
    tr[cnt].sum = (ll)a[x];
    for(unsigned int i = 0; i < G[x].size(); ++i)
    {
        int id = G[x][i];
        dfs(id);
    }
    sec[x] = ++cnt;
    tr[cnt].fl[0] = 1;
    tr[cnt].val = -a[x];
    tr[cnt].sum = (ll)(-a[x]);
}

void build(int l, int r, int ff)
{
    int mid = (l + r)>>1;
    if(l < mid)
        build(l, mid - 1, mid);
    if(mid < r)
        build(mid + 1, r, mid);
    if(ff)
        tr[ff].s[ff < mid] = mid;
    int ls = tr[mid].s[0], rs = tr[mid].s[1];
    tr[mid].fa = ff;
    tr[mid].cnt1 = tr[ls].cnt1 + tr[rs].cnt1 + tr[mid].fl[1];
    tr[mid].cnt_1 = tr[ls].cnt_1 + tr[rs].cnt_1 + tr[mid].fl[0];
    tr[mid].sum += tr[ls].sum + tr[rs].sum;
}

void spread(int x)
{
    if(tr[x].add)
    {
        int ls = tr[x].s[0], rs = tr[x].s[1];
        if(ls)
        {
            tr[ls].sum += 1LL * (tr[ls].cnt1 - tr[ls].cnt_1) * tr[x].add;
            tr[ls].val += (tr[ls].fl[1]? tr[x].add: -tr[x].add);
            tr[ls].add += tr[x].add;
        }
        if(rs)
        {
            tr[rs].sum += 1LL * (tr[rs].cnt1 - tr[rs].cnt_1) * tr[x].add;
            tr[rs].val += (tr[rs].fl[1]? tr[x].add: -tr[x].add);
            tr[rs].add += tr[x].add;
        }
        tr[x].add = 0;
    }
}

void update(int x)
{
    int ls = tr[x].s[0], rs = tr[x].s[1];
    tr[x].sum = tr[ls].sum + tr[rs].sum + 1LL * tr[x].val;
    tr[x].cnt1 = tr[ls].cnt1 + tr[rs].cnt1 + tr[x].fl[1];
    tr[x].cnt_1 = tr[ls].cnt_1 + tr[rs].cnt_1 + tr[x].fl[0];
}

void Rotate(int x)
{
    int y = tr[x].fa, z = tr[y].fa, k = (tr[y].s[1] == x), w = (tr[z].s[1] == y), son = tr[x].s[k^1];
    spread(y);spread(x);
    tr[y].s[k] = son;tr[son].fa = y;
    tr[x].s[k^1] = y;tr[y].fa = x;
    tr[z].s[w] = x;tr[x].fa = z;
    update(y);update(x);
}

void Splay(int x, int to)
{
    int y, z;
    while(tr[x].fa != to)
    {
        y = tr[x].fa;
        z = tr[y].fa;
        if(z != to)
            Rotate((tr[y].s[0] == x) ^ (tr[z].s[0] == y)? x: y);
        Rotate(x);
    }
    if(!to)
        root = x;
}

int Findpre()
{
    spread(root);
    int now = tr[root].s[0];
    while(tr[now].s[1])
    {
        spread(now);
        now = tr[now].s[1];
    }
    spread(now);
    return now;
}

int Findnxt()
{
    spread(root);
    int now = tr[root].s[1];
    while(tr[now].s[0])
    {
        spread(now);
        now = tr[now].s[0];
    }
    spread(now);
    return now;
}

void Remove(int x, int y)
{
    Splay(fir[x], 0);
    int pre = Findpre();
    Splay(sec[x], 0);
    int nxt = Findnxt();
    Splay(pre, 0);
    Splay(nxt, pre);
    int now = tr[nxt].s[0];
    tr[nxt].s[0] = 0;
    update(nxt);update(pre);
    Splay(fir[y], 0);
    nxt = Findnxt();
    tr[now].fa = nxt;
    tr[nxt].s[0] = now;
    Splay(now, 0);
}

int main()
{
    read(n);
    for(int i = 1; i < n; ++i)
    {
        int x;
        read(x);
        G[x].push_back(i+1);
    }
    for(int i = 1; i <= n; ++i)
        read(a[i]);
    cnt = 1;
    dfs(1);
    cnt ++;
    build(1, cnt, 0);
    tr[0].s[1] = root = (1 + cnt)>>1;
    read(m);
    for(int i = 1; i <= m; ++i)
    {
        char opt[3];
        scanf("%s", opt);
        if(opt[0] == 'Q')
        {
            int x;
            read(x);
            Splay(sec[x], 0);
            printf("%lld\n", tr[tr[root].s[0]].sum);
        }
        else if(opt[0] == 'C')
        {
            int x, y;
            read(x);read(y);
            Remove(x, y);
        }
        else
        {
            int x, y;
            read(x);read(y);
            if(!y)    continue;
            Splay(fir[x], 0);
            int pre = Findpre();
            Splay(sec[x], 0);
            int nxt = Findnxt();
            Splay(pre, 0);
            Splay(nxt, pre);
            int p = tr[nxt].s[0];
            tr[p].sum += 1LL * (tr[p].cnt1 - tr[p].cnt_1) * y;
            tr[p].val += (tr[p].fl[1]? y: -y);
            tr[p].add += y;
            update(nxt);update(pre);
        }
    }
    return 0;
}

标签：星系,rs,int,tr,mid,Splay,ls,now,Bzoj3786
来源： https://www.cnblogs.com/Joker-Yza/p/11366884.html