PAT A1051

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

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#include "stdio.h"
//#include "math.h"
#include "string.h"
//#include "iostream"
#include "stdlib.h"
//#include "vector"
//#include "set"
//#include "map"
#include "stack"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn = 1010;
int arr[maxn];//保存题目给定的出栈序列
stack<int> st;//定义栈st 用以存放int型元素
int main(){
int m, n, T;
scanf("%d%d%d", &m, &n, &T);
while (T--) {
while (!st.empty()) {//清空栈
st.pop();
}
for (int i = 1; i <= n; i++) {
scanf("%d", &arr[i]);
}
int current = 1;//指向出栈序列中的待出栈元素
bool flag = true;
for (int i = 1; i <= n; i++) {
st.push(i);//把i压入栈
if (st.size() > m) {//如果此时栈中元素大于容量m,则序列非法
flag = false;
break;
}
//栈顶元素与出栈序列当前位置的元素相同时
while (!st.empty() && st.top() == arr[current]) {
st.pop();//反复弹栈并令current++
current++;
}
}
if (st.empty() == true && flag == true) {
printf("YES\n");
}else{
printf("NO\n");
}
}

return 0;
}