Pop Sequence

参考：http://blog.csdn.net/xyt8023y/article/details/46967345

题目如下：

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
#include<iostream>#include<iomanip>#include<stack>#include<vector> using namespace std;int main(){stack<int> nums;int N,M,K;int i,j,x;int index=1,nextToPush=1;cin>>N>>M>>K;vector<int> vec(N);bool flag[K];for(i=0;i<K;i++){for(j=0;j<M;j++){cin>>vec[j];}index=nextToPush=1;while(!nums.empty())nums.pop();flag[i]=1;while(index<N){x=vec[index-1];index++;if(x<nextToPush){if(!nums.empty() && nums.top()==x)nums.pop();else{flag[i]=0;break;}}else if(x==nextToPush){nextToPush++;}else{if(nextToPush>N){//重点，判断是否超出栈的容量，带入第三个用例理解 flag[i]=0;break;}for(;nextToPush<=x;nextToPush++){nums.push(nextToPush);if(nums.size()>N){flag[i]=0;break;}}nums.pop();}}}for(i=0;i<K;i++){if(flag[i]) cout<<"Yes"<<endl;else cout<<"No"<<endl;}return 0;}