浙大数据结构题目

线性结构4 Pop Sequence（25 分）
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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

题目的意思是:
1. 有一个堆栈限制大小为M , 测试K组数据数列, 每组数列的长度是N;
2. 以1,2,3…N的顺序往队列压数据,可以随时pop处理;判断是否pop出来的数据的顺序和我们给的数列(长度也是N)是否相同;相同YES,不相同NO;
3. 一个测试K组:

以下是用C++完成

#include <stdio.h>#include <stdlib.h>#include <stdbool.h>#include <stack>#include <iostream>#include <algorithm>using namespace std; int main() {    int M;//Max of stack   5    int N;//length of push sequence 7    int K;//num of pop sequence to be check      cin>>M>>N>>K;     stack<int> sta;    int v;     int flag = true;     int tmp;     for(int i=0;i<K;i++)    {        flag = true;        tmp = 1;         for(int j =0;j<N;j++)           {            cin>>v; //一定要输完的             while(flag&&sta.size()<=M){                if(sta.size()==0||sta.top()!=v){//放入                     sta.push(tmp++);                } else if(sta.top()==v){//顶部相同                     sta.pop();//推出一个后应该指针 转到下一个                     break;                 }             }             if(sta.size()>M){                flag= false;              }            }         if(flag){            printf("YES\n");         }else{            printf("NO\n");         }         while(!sta.empty())              sta.pop();      }     return 0; }