02-线性结构4 Pop Sequence (25分)

Given a stack which can keep MMM numbers at most. Push NNN numbers in the order of 1, 2, 3, …, NNN 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 MMM is 5 and NNN 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): MMM (the maximum capacity of the stack), NNN (the length of push sequence), and KKK (the number of pop sequences to be checked). Then KKK lines follow, each contains a pop sequence of NNN 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<iostream>#include<stack>#include<cstdio>#include<vector>#include<algorithm>using namespace std;int M,N,K;int check(vector<int> &v){    int idx=0,num=1,cap=M+1;    stack<int> sta;    sta.push(0);    while(idx<N){//当v[idx]的值大于栈顶序列时说明它前面的值已经全部压入栈里了，如果此时栈还不满则继续入栈        while(v[idx]>sta.top()&&sta.size()<cap)            sta.push(num++);        if(v[idx++]==sta.top())     sta.pop();//如果v[idx]的值等于栈顶序列则pop出去这个元素，且idx++        else return 0;    }    return 1;}int main(){    vector<int> vec(N,0);//N为个数，0为初始值    scanf("%d%d%d",&M,&N,&K);    for(int i=0;i<K;i++){        for(int j=0; j<N; j++){            int temp;            scanf("%d",&temp);            vec.push_back(temp);        }        if(check(vec))  printf("YES\n");        else    printf("NO\n");        vec.clear();            //最后记得清空vector用于第二次输入    }    return 0;}