HDU 5818 Joint Stacks（联合栈）
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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Description |
题目描述 |
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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. |
栈是一种数据结构，仅允许在其中一端进行插入和删除，这一端被称为栈顶。最后入栈的元素将会最先出栈。换句话说，这个操作就是先进先出(LIFO)。 一个可合并的栈拥有"merge"操作。有如下三种操作： - push A x: 将 x 插入栈 A - pop A: 弹出 A 的栈顶元素 - merge A B: 合并栈 A 与 B 执行"merge A B"操作后，栈A将获得B中全部元素，随后B则为空。新栈中的元素根据先前的入栈时间重新排列，如同再执行他们的"push"操作到一个栈。参考输入/输出样例的详细说明。 给定两个可合并栈A与B，执行上述操作。 |
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Input |
输入 |
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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. |
多组测试用例。对于每个测试用例，第一行有一个整数N(0<N≤105)，表示操作的数量。随后N行，每行包含一个"push"、 "pop"或"merge"指令。栈中的元素都是32位整数。保证"pop"操作不会出现在空栈中。N = 0时输入结束。 |
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Output |
输出 |
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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. |
对于每个用例，输出一行"Case #t:"，t表示用例编号(从1开始)。对于每个"pop"操作，输出出栈的元素在单独一行。 |
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Sample Input - 输入样例 |
Sample Output - 输出样例 |
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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 |
Case #1: 2 1 Case #2: 1 2 3 0 Case #3: 1 2 3 0 |
【题解】
使用三个优先队列，一个给A，一个B，一个给合并的部分。
当A/B为空的时候，就从合并的部分取。
只用AB的话，感觉就不靠谱，懒癌发作，不想试。
【代码 C++】
1 #include <cstdio>
2 #include <queue>
3 #define mx 100005
4 int data[mx], id;
5 int main(){
6 int t = 1, n, x;
7 char op[10], o;
8 while (scanf("%d", &n), n){
9 printf("Case #%d:\n", t++); id = -1;
10 std::priority_queue<int, std::vector<int> > a, b, m;
11
12 while (n--){
13 scanf("%s", op); getchar();
14 if (op[1] == ‘u‘){
15 scanf("%c%d", &o, &data[++id]);
16 if (o == ‘A‘) a.push(id);
17 else b.push(id);
18 }
19 else if (op[1] == ‘o‘){
20 if (getchar() == ‘A‘){
21 if (a.empty()) x = data[m.top()], m.pop();
22 else x = data[a.top()], a.pop();
23 }
24 else{
25 if (b.empty()) x = data[m.top()], m.pop();
26 else x = data[b.top()], b.pop();
27 }
28 printf("%d\n", x);
29 }
30 else{
31 for (gets(op); !a.empty(); a.pop()) m.push(a.top());
32 while (!b.empty()) m.push(b.top()), b.pop();
33 }
34 }
35 }
36 return 0;
37 }