【详细注释】1051 Pop Sequence (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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

#include<bits/stdc++.h>
using namespace std;
int main(){int M,N,K;scanf("%d%d%d",&M,&N,&K);while(K--){int A[N],t=0;//数组A存储出栈顺序，t存储下一次压栈时的首元素-1stack<int>s;bool f=true;//判断输出YES或者NOfor(int i=0;i<N;++i)scanf("%d",&A[i]);for(int i=0;i<N&&f;++i){//遍历数组Aif(s.empty()||s.top()<A[i]){//栈为空或者栈顶元素小于A[i]for(int k=t+1;k<=A[i]&&s.size()<M;++k)//将t+1~A[i]这些数按从小到大顺序压入栈中，并保证栈的当前容量不超过Ms.push(k);t=s.empty()?0:s.top();//更新tif(t==A[i])//栈顶元素与A[i]相等s.pop();//弹出else//栈顶元素与A[i]不等f=false;//置f为false}else if(s.top()==A[i])//栈不为空且栈顶元素等于A[i]s.pop();//弹出栈顶元素else//栈不为空且栈顶元素大于A[i]f=false;//置f为false}printf("%s\n",f?"YES":"NO");//输出}return 0;
}

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【详细注释】1051 Pop Sequence (25 分)

立志用最少的代码做最高效的表达

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