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
按照其入栈队列模拟入栈出栈操作： 1.若栈顶元素x 小于 出栈队列中当前需要出栈的元素y，则将小于等于y的元素依次入栈 2.若栈顶元素x 大于 出栈队列中当前需要出栈的元素y，则说明y在堆栈中但是不在栈顶，根据堆栈后进先出的规则，只能栈顶元素出栈，所以该队列为不可能队列 3.若栈顶元素x 等于 出栈队列中当前需要出栈的元素y，则将栈顶元素出栈
#include<stdio.h> #include<iostream> #include<stdlib.h> #include<string.h> using namespace std; int stack[1002] = {0}; int top = 1;//栈顶指针 int p = 1;//进栈队列头 void push(){ stack[top++] = p++; } int pop(){ return stack[--top]; } bool isPossible(int a[], int M, int N){ for(int i = 0; i < N; i++){ if(a[i] > stack[top-1]){ for(int j = p; j <= a[i];j++) push(); if(top > M+1) return false; pop(); }else if(a[i] == stack[top-1]){ pop(); }else{ return false; } } return true; } int main(){ int M,N,K; cin >> M >> N >> K; while(K--){ int *a; a = (int*)malloc(N*sizeof(int)); for(int i = 0; i < N; i++){ cin >> a[i]; } top = 1; p = 1; for(int i = 0; i < M+1; i++) stack[i] = 0; if(isPossible(a,M,N)){ printf("YES\n"); }else{ printf("NO\n"); } } return 0; }