Problem Description
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.
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.
Input
There are multiple test cases. For each case, the first line contains an integer N(0<N≤105), indicating the number of operations. The next N lines, each contain an instruction "push", "pop" or "merge". The elements of stacks are 32-bit integers. Both A and B are empty initially, and it is guaranteed that "pop" operation would not be performed to an empty stack. N = 0 indicates the end of input.
Output
For each case, print a line "Case #t:", where t is the case number (starting from 1). For each "pop" operation, output the element that is popped, in a single line.
Sample Input
4
push A 1
push A 2
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge A B
pop A
pop A
pop A
9
push A 0
push A 1
push B 3
pop A
push A 2
merge B A
pop B
pop B
pop B
0
Sample Output
Case #1:
2
1
Case #2:
1
2
3
0
Case #3:
1
2
3
0
思路：用C栈保存A、B栈的合并，可以再用一个D栈保存A和B的合并，然后再导到C栈；
代码如下：
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <stack> #include <queue> using namespace std; struct Node { int x; int t; }; stack<Node*>A,B,C,D; int main() { int N,Case=1; char s[10]; char x; int c; while(scanf("%d",&N)&&N) { printf("Case #%d: ",Case++); for(int i=1;i<=N;i++) { scanf("%s",s); if(s[1]=='u') { scanf(" %c",&x); scanf("%d",&c); Node* a=new Node(); a->x=c; a->t=i; if(x=='A') A.push(a); else B.push(a); } else if(s[1]=='o') { scanf(" %c",&x); if(x=='A') { if(A.empty()) { printf("%d ",C.top()->x); C.pop(); } else { printf("%d ",A.top()->x); A.pop(); } } else { if(B.empty()) { printf("%d ",C.top()->x); C.pop(); } else { printf("%d ",B.top()->x); B.pop(); } } } else if(s[1]=='e') { scanf(" %c",&x); scanf(" %c",&x); while(!A.empty()&&!B.empty()) { if(A.top()->t>B.top()->t) { D.push(A.top()); A.pop(); } else { D.push(B.top()); B.pop(); } } while(!A.empty()) { D.push(A.top()); A.pop(); } while(!B.empty()) { D.push(B.top()); B.pop(); } while(!D.empty()) { C.push(D.top()); D.pop(); } } } } return 0; }