PAT 解题报告 1051. Pop Sequence (25)

 

1051. Pop Sequence (25)

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

题目描述：

一列数, 只能以1, 2, …., N的push 到stack里面, 但是可以再任意时刻pop出一个数字, 问给定一个序列, 是不是一个可行的按照这样的规则可以出现的一个pop序列.

算法分析：

思路：总结规律，使用hash

使用hash记录已经弹出栈的数字

从前向后扫，两种情况下是不可能的

1、比如当前扫到了7，查看1~6是否已经弹出，若没有弹出的数量count<M(栈的容量)，则说明不可能。

2、比如当前扫到了7，而前一个是4，就要检测5和6是否弹出，如果有一个没弹出，则不可能

注意点：

无

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>

using namespace std;

int hashs[1002];
int M, N, K;

int getScale(int x){
    int cnt = 0;
    for (int i=1; i<=x; i++) {
        if (hashs[i] == 0)
            cnt++;
    }
    return cnt;
}
bool ckinv(int x,int y){
    for (int i=y+1; i<x; i++){
        if (hashs[i] == 0)
            return false;
    }
    return true;
}

int main()
{
    scanf("%d%d%d", &M, &N, &K);
    for (int i=0; i<K; i++) {
        memset(hashs, 0, sizeof(hashs));
        int a[N+1];
        for (int j=1; j<=N;j++) {
            scanf("%d", a+j);
        }
        int pre;
        bool flag = true;
        for (int j=1; j<=N; j++) {
            int tmp = a[j];
            if (getScale(tmp) > M) {
                printf("NO
");
                flag = false;
                break;
            }
            if (j!=1 && ckinv(pre, tmp)== false) {
                printf("NO
");
                flag = false;
                break;
            }
            hashs[tmp] = 1;
            pre = tmp;
        }
        if (flag == true) printf("YES
");
    }
    return 0;
}