1051 Pop Sequence (25分)

题目

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 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

题目大意

模拟栈操作，不过这个栈的大小有限制，输入序列限定为$1\to N$，通过模拟出栈入栈操作，看能否得到预期输出序列。

代码

#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;

int main(){
    int m, n, k;
    int a[1001];
    stack<int> sta;
    scanf("%d%d%d", &m, &n, &k);
    while(k--){
        for(int i=0; i<n; i++)
            scanf("%d", a+i);
        int t = 1, curi = 0;
        while(t <= n){
            if(sta.empty() || sta.top() != a[curi]){
                if(sta.size() == m)
                    break;
                sta.push(t);
                //printf("入栈%d\n", t);
                t++;
            }
            else{
               // printf("出栈%d\n", sta.top());
                sta.pop();
                curi++;
            }
        }
        while(!sta.empty() && sta.top() == a[curi]){
            //printf("出栈%d\n", sta.top());
            sta.pop();
            curi++;
        }
        if(sta.empty())
            printf("YES\n");
        else
            printf("NO\n");
        while(!sta.empty())
            sta.pop();
    }
    return 0;
}