#### HDU 5818 – Joint Stacks [左偏树/优先队列]

http://acm.hdu.edu.cn/showproblem.php?pid=5818

### Problem

A stack is a data structure in which all insertions and deletions of entries are made at one end, called the “top” of the stack. The last entry which is inserted is the first one that will be removed. In another word, the operations perform in a Last-In-First-Out (LIFO) manner.
A mergeable stack is a stack with “merge” operation. There are three kinds of operation as follows:

– push A x: insert x into stack A
– pop A: remove the top element of stack A
– merge A B: merge stack A and B

After an operation “merge A B”, stack A will obtain all elements that A and B contained before, and B will become empty. The elements in the new stack are rearranged according to the time when they were pushed, just like repeating their “push” operations in one stack. See the sample input/output for further explanation.
Given two mergeable stacks A and B, implement operations mentioned above.

### Code1

```#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;

const int maxn = 152400;

P v[maxn];
int tot, u, a, b,  l[maxn], r[maxn], d[maxn];

int Merge(int x, int y) {
if(!x) return y;
if(!y) return x;
if(v[x] < v[y]) swap(x, y);
r[x] = Merge(r[x], y);
if(d[l[x]] < d[r[x]]) swap(l[x], r[x]);
d[x] = d[r[x]] + 1;
return x;
}
int init(P x) {
tot++;
v[tot] = x;
l[tot] = r[tot] = d[tot] = 0;
}
int Insert(int x, P y) {
return Merge(x, init(y));
}
P top(int x) {
return v[x];
}
int pop(int x) {
return Merge(l[x], r[x]);
}

char op[30], str[30], str2[30];
int val;

int main() {
int kase = 0;
int N;
while(scanf("%d", &N), N) {
memset(v, 0, sizeof(v));
memset(l, 0, sizeof(l));
memset(r, 0, sizeof(r));
memset(d, 0, sizeof(d));
tot = u = a = b = 0;

printf("Case #%d:\n", ++kase);
int posA = init(P(-1, -1));
int posB = init(P(-2, -1));

for(int i = 1; i <= N; ++i) {
scanf("%s%s", op, str);
if(op[1] == 'u') {
scanf("%d", &val);
if(str[0] == 'A') {
posA = Insert(posA, P(i, val));
} else {
posB = Insert(posB, P(i, val));
}
}
else if(op[1] == 'o') {
P ans;
if(str[0] == 'A') {
ans = top(posA);
posA = pop(posA);
} else {
ans = top(posB);
posB = pop(posB);
}
printf("%d\n", ans.second);
}
else if(op[1] == 'e') {
scanf("%s", str2);
if(str[0] == 'A') {
posA = Merge(posA, posB);
posB = init(P(-i, -1));
} else {
posB = Merge(posA, posB);
posA = init(P(-i, -1));
}
}
}
}
return 0;
}
```

### Code2

```#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;

char op[30], str[30], str2[30];

int main() {
int N, kase = 0;
while(scanf("%d", &N), N) {
priority_queue<P> A, B, C;
printf("Case #%d:\n", ++kase);
for(int i = 1; i <= N; ++i) {
scanf("%s%s", op, str);
if(op[1] == 'u') {
int val; scanf("%d", &val);
if(str[0] == 'A') {
A.push(P(i, val));
} else {
B.push(P(i, val));
}
}
else if(op[1] == 'o') {
P ans;
if(str[0] == 'A') {
if(!A.empty()) ans = A.top(), A.pop();
else ans = C.top(), C.pop();
} else {
if(!B.empty()) ans = B.top(), B.pop();
else ans = C.top(), C.pop();
}
printf("%d\n", ans.second);
}
else if(op[1] == 'e') {
scanf("%s", str2);
while(!A.empty()) {
C.push(A.top());
A.pop();
}
while(!B.empty()) {
C.push(B.top());
B.pop();
}
}
}
}
return 0;
}
```