1051. Pop Sequence (25)

题目：

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

解答：

//A1051 #include<cstdio>#include<stack>using namespace std;int main(){    int m,n,k;    scanf("%d%d%d",&m,&n,&k);    stack<int> st;    int arr[1010];     while(k--){//k在后文不回再用到，可以用这种方式减少循环代码量         while(!st.empty())//每次输入前均清原有空栈             st.pop();        for(int i=1;i<=n;i++)//读入数据             scanf("%d",&arr[i]);        int current=1;        for(int i=1;i<=n;i++){            st.push(i);            if(st.size()>m) break;            //栈顶元素和当前arr要出列元素相同时不断弹出，current指针向后移             while(!st.empty()&&st.top()==arr[current]){                st.pop();                current++;            }        }        if(current==n+1)printf("YES\n");//如果按顺序读完了所有的元素，说明没问题         else printf("NO\n");    }    return 0;}