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### stostap's blog

By stostap, history, 3 years ago, , Hi community,

I'm big fan of treaps, but this task makes me feel stupid. https://www.e-olymp.com/en/problems/2957

There is no English translate for some reason. Task: Given N single-element lists with integers 1..10^9, perform next queries:

• merge L R -> take two already exist lists and create a new one, equal concat(L,R)
• head L -> create two new lists: first contains first element of L, second — the rest of L.
• tail L -> create two new lists: first contains all L without last element, second — last element.

For every new list you need to output sum of it's elements.

I've did it with persistent treap — for every merge query: merge(root[L], root[R], root[++versions]), for head — split(root[L], root[nextV()], root[nextV()], 1) for tail — split(root[L], root[nextV()], root[nextV()], cnt[root[L]]-1);

But this fails with "Memory Limit" as for every query it creates at least Log new nodes

typedef long long LL;
#define N 13000005
#define MOD 1000000007
int root , c;
struct Treap
{
int sum[N];
int nodecnt;
int L[N] , R[N] , cnt[N];
int key[N];
void clear() {
nodecnt = 0;
}
Treap () {clear();}
bool hey(int A , int B) {
return (LL)rand() * (cnt[A] + cnt[B]) < (LL)cnt[A] * RAND_MAX;
}
int newnode(int x) {
++ nodecnt , L[nodecnt] = R[nodecnt] = 0;
cnt[nodecnt] = 1 , key[nodecnt] = x, sum[nodecnt] = x;
return nodecnt;
}
int copynode(int A) {
if (!A) return 0;
++ nodecnt , L[nodecnt] = L[A] , R[nodecnt] = R[A];
cnt[nodecnt] = cnt[A] , key[nodecnt] = key[A], sum[nodecnt] = sum[A];
if (nodecnt == 5000000 &mdash; 100) {
printf("TREAP\n");
exit(0);
}
return nodecnt;
}
void pushup(int x) {
cnt[x] = 1;
sum[x] = key[x];
if (L[x]) {
cnt[x] += cnt[L[x]];
sum[x] = (sum[x] + sum[L[x]]) % MOD;
}
if (R[x]) {
cnt[x] += cnt[R[x]];
sum[x] = (sum[x] + sum[R[x]]) % MOD;
}
}
void merge(int& p , int x , int y) {
if (!x || !y) {
p = 0;
if (x) p = copynode(x);
if (y) p = copynode(y);
}
else if ( hey(x , y) ) {
p = copynode(x);
merge(R[p] , R[x] , y) , pushup(p);
}
else {
p = copynode(y);
merge(L[p] , x , L[y]) , pushup(p);
}
}
void split(int p , int& x , int& y , int size) {
if (!size) {
x = 0 , y = copynode(p);
return;
}
if (cnt[L[p]] >= size) {
y = copynode(p);
split(L[p] , x , L[y] , size) , pushup(y);
}
else {
x = copynode(p);
split(R[p] , R[x] , y , size &mdash; cnt[L[p]] &mdash; 1) , pushup(x);
}
}
void print(int p) {
if (L[p]) print(L[p]);
printf("%d ", key[p]);
if (R[p]) print(R[p]);
}
};
Treap T;
char s;

int main(void) {
int n;
scanf("%d", &n);
int version = 0;
REP(i, n) {
int x;
scanf("%d", &x);
root[++version] = T.newnode(x);
}

int q;
scanf("%d", &q);
REP(i, q) {
if (version >= 300005 &mdash; 50) {
printf("TREAP2\n");
exit(0);
}
scanf("%s", &s);
if (s == 'm') {
int x,y;
scanf("%d%d", &x, &y);
T.merge(root[++version], root[x], root[y]);
printf("%d\n", T.sum[root[version]]);
}

if (s == 'h') {
int x;
scanf("%d", &x);
int v1 = ++version;
int v2 = ++version;
T.split(root[x], root[v1], root[v2], 1);
printf("%d\n%d\n", T.sum[root[v1]], T.sum[root[v2]]);
}

if (s == 't') {
int x;
scanf("%d", &x);
int v1 = ++version;
int v2 = ++version;
T.split(root[x], root[v1], root[v2], T.cnt[root[x]] - 1);
printf("%d\n%d\n", T.sum[root[v1]], T.sum[root[v2]]);
}
}
}


Any ideas how to improve it?  Comments (13)
 » Auto comment: topic has been updated by stostap (previous revision, new revision, compare).
 » 3 years ago, # | ← Rev. 5 →   UPD: sorry i got the problem wrong.scarped until I get a new solution :P
•  » » As far as I understood when you do head/tail operation you also need to keep the previous list and then add two more lists. How do you handle this?
 » I didn't get what you meant by "But this fails as for every query it creates at least Log new nodes". In fact you cannot do a persistent treap in complexity better than , because for each merge or split you create new nodes.
 » Auto comment: topic has been updated by stostap (previous revision, new revision, compare).
 » Auto comment: topic has been updated by stostap (previous revision, new revision, compare).
 » What do you mean by "it fails"? Wrong Answer, Time Limit Exceeded, Memory Limit Exceeded, Runtime Error? Maybe something even more specific?
•  » » Sorry, yes, it's memory limit exceeded in few tests. More here: https://www.e-olymp.com/uk/submissions/3233986I have Runtime Error here, but I explicitly calling abort(), when we are close to ML.
 » Auto comment: topic has been updated by stostap (previous revision, new revision, compare).
 » I believe it's near to impossible to solve this problem with a simple persistent treap. Trap here is that depth of the tree is , where n is the number of all elements in the tree, which can be doubled with a single operation. Consider the following test case: There is only a single list initially. One performs 5·104 merge operations, each of which doubles the largest list. This results in a list with 25·104 elements, e.g. a tree of depth  ≈ 5·104 (yes, I know that writing such big-O is meaningless from asymptotical analysis' point of view, but I believe my idea should be understandable). We perform 5·104 head operations on that largest list, each operation creates  ≈ 5·104 new nodes. That sums up to  ≈ 2.5·109 new nodes which is both TL and ML regardless of any implementation details.Hint: note that you actually cannot split the list at arbitrary point, you can only chop few of its first/last elements. Then create a smarter approach (which will still use persistent treap, though).
•  » » Yes, I know that depth is a problem. My main observation is that we could "cache" some vertex, instead of creating a new one.Obvious one is a vertex with left or right result of head/tail. Now thinking about what will be appropriate "hash" for a vertex in treap.
•  » » » 3 years ago, # ^ | ← Rev. 2 →   Storing just one vertex doesn't help you as you cannot recalculate that when cutting head/tail off.
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   hm, I just crazy idea not performing actual splitting for head/tail, but just have 2 counters — cnt_left_cur & cnt_right_cut. When head/tail query comes — just make 1 split & 1 merge without creating new nodes (in-place) in actual tree — this shouldn't change a tree. When merge comes — make 2 splits (from head and tail) and merge results.This is trick for head/tail queries. But I can't find anything for merge :(UPD:OK, for merge we could not merge them actually, but just hold that this treap constist of T1 & T2. Then when cnt_left_cut > T1.cnt totally forgot about T1, and working only with pair(T2, 0). Similar for cnt_right_cur