1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

题意：给定一个栈的容量，然后将1-N push进去，给定一个1-N的全排列，问能否经过一系列操作得出这个序列

题解：模拟栈的操作

#include <cstdio>#include <stack>#include <queue>using namespace std;int main() {    int m, n, k;    scanf("%d%d%d", &m, &n, &k);    while (k--) {        stack<int> s;        queue<int> check;        bool flag = true;        for (int i = 1; i <= n; ++i) {            int tmp;            scanf("%d", &tmp);            check.push(tmp);            s.push(i);            if (s.size() > m) flag = false;            while (!s.empty() && s.top() == check.front()) {                s.pop();                check.pop();            }        }        if (!s.empty()) flag = false;        if (flag == true) puts("YES");        else puts("NO");    }    return 0;}