树链剖分

重链剖分

int dep[Z], siz[Z], dad[Z], kid[Z];//kid[rt]：rt的重儿子编号
int dfn[Z], tim, top[Z], rnk[Z];//top[rt]：rt所在重链的顶部节点; rnk[rt]：dfs序为rt的节点编号
void search(int rt, int fa)//寻找重边和重儿子
{
    dad[rt] = fa, kid[rt] = 0;
    dep[rt] = dep[fa] + 1, siz[rt] = 1;
    int mxsz = 0;
    for (re i = head[rt]; i; i = e[i].ne)
    {
        int son = e[i].to;
        if (dep[son]) continue;
        search(son, rt);
        siz[rt] += siz[son];
        if (mxsz < siz[son]) mxsz = siz[son], kid[rt] = son;//更新重儿子
    }
}
void connect(int rt, int pa)//连接重边构成重链
{
    dfn[rt] = ++tim; rnk[tim] = rt;
    top[rt] = pa;
    if (kid[rt]) connect(kid[rt], pa);//优先递归重儿子
    for (re i = head[rt]; i; i = e[i].ne)
    {
        int son = e[i].to;
        if (son == kid[rt] || son == dad[rt]) continue;
        connect(son, son);//新建立一条重链
    }
}
void modify(int x, int y, int val)//对从x到y的简单路径上的点权值加val
{
    while (top[x] != top[y])//没在一条重链上
    {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        update(dfn[top[x]], dfn[x], val);
        x = top[x];//向上翻
    }
    if (dep[x] > dep[y]) swap(x, y);
    update(dfn[x], dfn[y], val);
}
int ask(int x, int y)//查询从x到y的简单路径上的点权值之和
{
    while (top[x] != top[y])//没在一条重链上
    {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += query(dfn[top[x]], dfn[x]);
        x = top[x];//向上翻
    }
    if (dep[x] < dep[y]) swap(x, y);
    ans += query(dfn[x], dfn[y]);
    return ans;
}

标签：rt,剖分,int,top,son,dep,dfn,树链
来源： https://www.cnblogs.com/zsj11337/p/16368369.html