pat advanced 1051

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1051

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

思路：

可以模拟出其出栈序列的抽象的规律：可以向后跳跃移动，但是只能连续向前移动。当然，隐含的条件是不能超出栈容量。

用程序来模拟的话，可以这样：

用指针index指代当前还未被push进站的元素的位置，如果当前栈为空，push一个元素，index自增，如果当前栈元素小于出栈序列的当前元素，一直push到相等(index自增)，否则如果当前元素小于出栈序列的当前元素，输出NO，等于的话，continue下一次判断。

参考代码：

#include<cstdio>#include<stack>using namespace std;int main(){    int M, N, K, arr[1001];    scanf("%d %d %d", &M, &N, &K);    for(int i = 0; i < K; ++i)    {        for(int j = 1; j <= N; ++j)        {            scanf("%d", arr+j);        }        stack<int> st;        bool is_yes = true;        int index = 1;        for(int j = 1; j <= N && is_yes; ++j)        {            if(st.empty())            {                st.push(index++);            }            if(st.top() < arr[j])            {                while(st.top() < arr[j])                    st.push(index++);                if(st.size() > M)                {                    is_yes = false;                    break;                }            }            if(st.top() != arr[j])            {                is_yes = false;                break;            }            st.pop();        }        is_yes ? printf("YES\n") : printf("NO\n");    }    return 0;}