SPOJ 2798 Query on a tree again!

SPOJ_2798

    如果用link-cut-tree写的话，只要维护col（节点的颜色）和sum（子树中black节点的数量）两个标记即可。染色的时候将对应节点splay到根然后修改，查询的时候先进行access(v)的操作，之后找到这棵splay的根再递归查找即可，如果根结点的sum值为0，则输出-1。

#include<stdio.h>
#include<string.h>
#define MAXD 100010
#define MAXM 200010
int N, Q, q[MAXD], first[MAXD], e, next[MAXM], v[MAXM];
struct Splay
{
    int pre, ls, rs, col, sum;
    bool root;
    void update(); void zig(int ); void zag(int ); void splay(int );
    void renew()
    {
        root = true;
        pre = ls = rs = 0;
        col = 0, sum = 0;
    }
}sp[MAXD];
void Splay::update()
{
    sum = sp[ls].sum + sp[rs].sum + col;
}
void Splay::zig(int x)
{
    int y = rs, fa = pre;
    rs = sp[y].ls, sp[rs].pre = x;
    sp[y].ls = x, pre = y;
    sp[y].pre = fa;
    if(root)
        root = false, sp[y].root = true;
    else
        sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y;
    update();
}
void Splay::zag(int x)
{
    int y = ls, fa = pre;
    ls = sp[y].rs, sp[ls].pre = x;
    sp[y].rs = x, pre = y;
    sp[y].pre = fa;
    if(root)
        root = false, sp[y].root = true;
    else
        sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y;
    update();
}
void Splay::splay(int x)
{
    int y, z;
    for(; !root;)
    {
        y = pre;
        if(sp[y].root)
            sp[y].rs == x ? sp[y].zig(y) : sp[y].zag(y);
        else
        {
            z = sp[y].pre;
            if(sp[z].rs == y)
            {
                if(sp[y].rs == x)
                    sp[z].zig(z), sp[y].zig(y);
                else
                    sp[y].zag(y), sp[z].zig(z);
            }
            else
            {
                if(sp[y].ls == x)
                    sp[z].zag(z), sp[y].zag(y);
                else
                    sp[y].zig(y), sp[z].zag(z);
            }
        }
    }
    update();
}
void add(int x, int y)
{
    v[e] = y;
    next[e] = first[x], first[x] = e ++;
}
void prepare()
{
    int i, j, x, rear = 0;
    sp[0].sum = 0;
    q[rear ++] = 1;
    sp[1].renew();
    for(i = 0; i < rear; i ++)
    {
        x = q[i];
        for(j = first[x]; j != -1; j = next[j])
            if(v[j] != sp[x].pre)
            {
                sp[v[j]].renew(), sp[v[j]].pre = x;
                q[rear ++] = v[j];
            }
    }
}
void init()
{
    int i, x, y;
    memset(first, -1, sizeof(first));
    e = 0;
    for(i = 1; i < N; i ++)
    {
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    prepare();
}
void change(int x)
{
    sp[x].splay(x);
    sp[x].col ^= 1;
    sp[x].update();
}
void access(int x)
{
    int fx;
    for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre)
    {
        sp[fx].splay(fx);
        sp[sp[fx].rs].root = true;
        sp[fx].rs = x, sp[x].root = false;
        sp[fx].update();
    }
}
int Search(int cur)
{
    if(sp[sp[cur].ls].sum != 0)
        return Search(sp[cur].ls);
    else if(sp[cur].col)
        return cur;
    else
        return Search(sp[cur].rs);
}
void ask(int x)
{
    int y;
    access(x);
    for(y = x; !sp[y].root; y = sp[y].pre);
    if(sp[y].sum == 0)
        printf("-1\n");
    else
        printf("%d\n", Search(y));
}
void solve()
{
    int i, op, x;
    for(i = 0; i < Q; i ++)
    {
        scanf("%d%d", &op, &x);
        if(op == 0)
            change(x);
        else
            ask(x);
    }
}
int main()
{
    while(scanf("%d%d", &N, &Q) == 2)
    {
        init();
        solve();
    }
    return 0;
}