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第18届浙江大学校赛 Mergeable Stack

發(fā)布時間：2025/3/15 编程问答 9 豆豆

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The 18th Zhejiang University Programming Contest Sponsored by TuSimple

第18屆浙江大學(xué)校賽的c題

解析：起先是用stack寫的模擬，但是先是爆內(nèi)存，然后在爆時間，結(jié)束后才知道，原來Stl在每次變長內(nèi)存空間的時候

是2倍的指數(shù)增長的，所以就是他的一個測試的樣例就是剛好會使stl兩倍的增長。時間爆是stack每次替換元素會o（n）。

這時候通常就是要手寫模擬了，開數(shù)組模擬。但是有一個東西特別，list，這個stl是指針的有關(guān)操作，就是很內(nèi)存還有時間就是

比較省，一步一步的增加的，時間也是，不會和stack指數(shù)增長。以下是代碼：（list的相關(guān)函數(shù)晚上）

152 - The 18th Zhejiang University Programming Contest Sponsored by TuSimple - CMergeable Stack

Time Limit:?2 Seconds ?????Memory Limit:?65536 KB

Given??initially empty stacks, there are three types of operations:

1?s?v: Push the value??onto the top of the?-th stack.
2?s: Pop the topmost value out of the?-th stack, and print that value. If the?-th stack is empty, pop nothing and print "EMPTY" (without quotes) instead.
3?s?t: Move every element in the?-th stack onto the top of the?-th stack in order.
Precisely speaking, denote the original size of the?-th stack by?, and the original size of the?-th stack by?. Denote the original elements in the?-th stack from bottom to top by?, and the original elements in the?-th stack from bottom to top by?.
After this operation, the?-th stack is emptied, and the elements in the?-th stack from bottom to top becomes?. Of course, if?, this operation actually does nothing.

There are??operations in total. Please finish these operations in the input order and print the answer for every operation of the second type.

Input

There are multiple test cases. The first line of the input contains an integer?, indicating the number of test cases. For each test case:

The first line contains two integers??and??(), indicating the number of stacks and the number of operations.

The first integer of the following??lines will be??(), indicating the type of operation.

If?, two integers??and??(,?) follow, indicating an operation of the first type.
If?, one integer??() follows, indicating an operation of the second type.
If?, two integers??and??(,?) follow, indicating an operation of the third type.

It's guaranteed that neither the sum of??nor the sum of??over all test cases will exceed?.

Output

For each operation of the second type output one line, indicating the answer.

Sample Input

2 2 15 1 1 10 1 1 11 1 2 12 1 2 13 3 1 2 1 2 14 2 1 2 1 2 1 2 1 2 1 3 2 1 2 2 2 2 2 2 3 7 3 1 2 3 1 3 3 2 1 2 1 2 2 2 3 2 3

Sample Output

13 12 11 10 EMPTY 14 EMPTY EMPTY EMPTY EMPTY EMPTY EMPTY

Submit????Status

#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<set> #include<vector> #include<stack> #include<map> #include<queue> #include<list> #include<algorithm> using namespace std;const int maxn=300005; list<int> l[maxn];int main() {int n,q,t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&q);for(int i=1; i<n+1; i++)l[i].clear();while(q--){int op;scanf("%d",&op);if(op==1){int s,v;scanf("%d%d",&s,&v);l[s].push_back(v);}if(op==2){int s;scanf("%d",&s);if(l[s].empty())puts("EMPTY");else{printf("%d\n",l[s].back());l[s].pop_back();}}if(op==3){int s,t;scanf("%d%d",&s,&t);l[s].splice(l[s].end(),l[t]);}}}return 0;}

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