A1051. Pop Sequence
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<cstdio> #include<iostream> #include<algorithm> #include<stack> using namespace std; int main(){ int N, M, K, temp; stack<int> stk; scanf("%d%d%d", &M, &N, &K); for(int i = 0; i < K; i++){ int index = 1, find = 1; for(int j = 0; j < N; j++){ scanf("%d", &temp); if(stk.empty()){ stk.push(index++); } if(stk.top() < temp){ while(stk.top() < temp && stk.size() < M){ stk.push(index++); } if(stk.top() < temp) find = 0; } if(stk.top() > temp){ find = 0; continue; } if(stk.top() == temp){ stk.pop(); continue; } } if(find == 0) printf("NO\n"); else printf("YES\n"); while(!stk.empty()){ stk.pop(); } } cin >> N; return 0; }
总结:
1、栈的模拟题。注意栈的大小有限制。
2、注意每次使用完毕后清空stack。