标签:1051 sequence int NO pop Pop stack 25 numbers
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
题解:使用栈,每压入一个元素就判断是否与当前输出序列值相等,相等就循环弹出,注意压栈前判满和弹出时判空
#include<bits/stdc++.h> using namespace std; const int maxn=1010; #define inf 0x3fffffff vector<int> ve[maxn]; int main(){ int m,n,k,temp; int A[maxn]; scanf("%d %d %d",&m,&n,&k); for(int i=0;i<k;i++){ for(int j=0;j<n;j++){ scanf("%d",&temp); ve[i].push_back(temp); } stack<int> s; int r=0; bool flag=true; for(int t=1;t<=n;t++){ if(s.size()<m){ s.push(t); } else{ flag=false; } while(!s.empty()&&s.top()==ve[i][r]){ r++; s.pop(); } } if(!s.empty()||r<n){ printf("NO\n"); } else{ printf("YES\n"); } } return 0; }
标签:1051,sequence,int,NO,pop,Pop,stack,25,numbers 来源: https://www.cnblogs.com/dreamzj/p/15022279.html
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