PAT A1057. Stack (30)

1057. Stack (30)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if N is odd.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<= 10⁵). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian

where key is a positive integer no more than 10⁵.

Output Specification:

For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print "Invalid" instead.

Sample Input:

17PopPeekMedianPush 3PeekMedianPush 2PeekMedianPush 1PeekMedianPopPopPush 5Push 4PeekMedianPopPopPopPop

Sample Output:

InvalidInvalid322124453Invalid

思路分析：

用一个栈模拟实现栈的输入输出即可，关键在于如何查询中值。可以利用分块思想处理，先确定待查询的数在哪一块儿，再在块儿内进行查询即可。

题解：

#include <cstdio>#include <stack>#include <cstring>using namespace std;const int MAX = 100001;const int MAXN = 317;int n, block[MAXN] = {0}, table[MAX] = {0};char order[11];stack<int> s;bool cmp(int a, int b){return a < b;}int main(){scanf("%d", &n);while(n--){scanf("%s", order);if(strcmp(order, "Pop") == 0){if(!s.empty()){int k = s.top();printf("%d\n", k);s.pop();block[k/(MAXN-1)]--;table[k]--;}else printf("Invalid\n");}else if(strcmp(order, "PeekMedian") == 0){if(s.empty()){printf("Invalid\n");continue;}int k, idx = 0, index = 0;if(s.size() % 2 == 0) k = s.size()/2;else k = (s.size()+1)/2;int sum = 0;while(sum + block[idx] < k){sum += block[idx++];}index = idx * (MAXN-1);while(sum + table[index] < k){sum += table[index++];}printf("%d\n", index);}else{int k;scanf("%d", &k); s.push(k);table[k]++;block[k/(MAXN-1)]++;}}return 0;}