【NOI集训】【XJ】可持久化左偏树

http://hzxjhs.com:83/contest/456

果断可并堆

#include 
#include 
#include 
#include 
#include 
#include 
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)
#define p t[z]
#define u t[x]
#define v t[y]
using namespace std;
typedef long long LL;
const int N = 200005;
struct arr { int d, w, l, r; } t[N * 100];
int n, q, a[N], sz, nz, ro[N];
int Merge(int x, int y) {
    if (!x || !y) return x + y;
    if (v.w > u.w) swap(x, y);
    int z = ++ sz; p = u;
    p.r = Merge(p.r, y);
    if (t[p.r].d > t[p.l].d) swap(p.r, p.l);
    p.d = t[p.r].d + 1;
    return z;
}
int newd(int y) { int x = ++ sz; u.w = y; return x; }
int Qry(int x, int y) {
    Rep(i, 1, y - 1) {
        x = Merge(u.l, u.r);
    }
    return u.w;
}
int main()
{
    scanf ("%d%d", &n, &q);
    Rep(i, 1, n) scanf ("%d", &a[i]), ro[i] = newd(a[i]);
    nz = n;
    Rep(i, 1, q) {
        int ty, x, y, z;
        scanf ("%d", &ty);
        if (ty == 1) {
            scanf ("%d%d%d", &z, &x, &y);
            ro[++ nz] = Merge(newd(z), Merge(ro[x], ro[y])); // c1[nz] = x, c2[nz] = y;
        } else if (ty == 2) {
            scanf ("%d", &x);
            // Merge()
        } else {
            scanf ("%d%d", &x, &y);
            printf ("%d\n", Qry(ro[x], y));
        }
    }
 
    return 0;
}