splay模板(BZOJ3224)

用splay实现二叉搜索树的模板，支持插入，删除，找前缀后缀，x的排名以及第x名的数。

#include <cstdio>
#define l(x) t[x].s[0]
#define r(x) t[x].s[1]
#define f(x) t[x].p
#define lc(x) (r(f(x)) == x)
#define st(a,b,c) t[a].s[c] = b; f(b) = a

const int N = 100005;
int m,x,op,tt,rt;
struct nd {int v,p,sz,s[2];}t[N];

void pu(int x) {t[x].sz = t[l(x)].sz+t[r(x)].sz+1;}
void rot(int x) {
    int y = f(x), z = f(y), lx = lc(x), ly = lc(y);
    st(y,t[x].s[!lx],lx); st(z,x,ly); st(x,y,!lx); pu(y);
}
void sp(int x) {
    for(int y; y = f(x); rot(x)) if(f(y)) {
        if(lc(x)^lc(y)) rot(x); else rot(y);
    }
    pu(x), rt = x;
}
void in(int x) {
    int c = rt;
    while(t[c].s[t[c].v<x]) t[c].sz++, c = t[c].s[t[c].v<x];
    if(c) t[c].sz++; t[c].s[t[c].v<x] = ++tt, f(tt) = c, t[tt].v = x, t[tt].sz = 1, sp(tt);
}
void del(int x) {
    int c = rt;
    while(1) {
        if(t[c].v == x) {x = c; break;}
        if(t[c].v < x) c = r(c); else c = l(c);
    } 
    sp(x);
    if(!l(x)) f(r(x)) = 0, rt = r(x);
    else if(!r(x)) f(l(x)) = 0, rt = l(x);
    else {
        f(l(x)) = 0, c = l(x);
        while(r(c)) c = r(c);
        sp(c), r(c) = r(x), f(r(x)) = c;
    }
}
int pre(int x) {
    int c = rt, ans = 0xcfcfcfcf;
    while(c) if(t[c].v < x) ans = t[c].v, c = r(c); else c = l(c);
    return ans;
}
int nxt(int x) {
    int c = rt, ans = 0x3f3f3f3f;
    while(c) if(t[c].v > x) ans = t[c].v, c = l(c); else c = r(c);
    return ans;
}
int rk(int x) {
    int c = rt, k = 0, ans = 0;
    while(c) if(t[c].v < x) ans = k+t[l(c)].sz+1, k += t[l(c)].sz+1, c = r(c); else c = l(c);
    return ans+1;
}
int sa(int x, int k) {
    if(t[l(x)].sz == k-1) return t[x].v;
    if(t[l(x)].sz >= k) return sa(l(x), k);
    return sa(r(x), k-t[l(x)].sz-1);
}

int main() {
    scanf("%d", &m);
    while(m--) {
        scanf("%d%d", &op, &x);
        if(op == 1) in(x);
        else if(op == 2) del(x);
        else if(op == 3) printf("%d\n", rk(x));
        else if(op == 4) printf("%d\n", sa(rt,x));
        else if(op == 5) printf("%d\n", pre(x));
        else printf("%d\n", nxt(x));
    }
    return 0;
}