02-线性结构4. Pop Sequence (25)

原题链接：http://www.patest.cn/contests/mooc-ds/02-%E7%BA%BF%E6%80%A7%E7%BB%93%E6%9E%844

02-线性结构4. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <stdio.h>
#include <stdlib.h>
#define MaxSize 1000
typedef struct StackRecord{
	int capacity;
	int top;
	int data[MaxSize];
}*Stack;
Stack CreateStack(int capacity) {
	Stack S = (Stack)malloc(sizeof(struct StackRecord));
	S->top = -1;
	S->capacity = capacity;
	return S;
}
int Push(Stack S, int x) {	//将x元素入栈，如果溢出则入栈失败返回0
	if(S->capacity - S->top <= 1)
		return 0;
	S->data[++S->top] = x;
	return 1;
}
int Top(Stack S) {	//返回栈顶元素，空栈时返回-1
	if(S->top >= 0)
		return S->data[S->top];
	else
		return -1;
}
void Pop(Stack S) {	//弹出栈顶元素
	S->top--;
}
void DisposeStack(Stack S) {
	free(S);
}
//模拟进栈出栈过程：依次入栈并同时与出栈序列的第一个元素对比；若相等则弹出栈顶元素，并消去出栈序列的首元素；
//全部已入栈后出栈序列中的元素全部被消去则返回1，否则返回0；
int IsPopSeq(int* popOrder, int capacity, int n) {
	Stack S = CreateStack(capacity);
	int head = 0;		//维护一个下标，指向出栈序列中还没被消去的第一个元素
	for(int node = 1; node <= n; node++) {	//入栈节点从1到k
		if(!Push(S, node)){		//如果入栈失败表示栈满，则返回0
			DisposeStack(S);
			return 0;
		}
		while(Top(S) == popOrder[head]) {
			Pop(S);
			head++;
		}
	}
	DisposeStack(S);
	if(head != n)		//出栈序列不为空，则返回0
		return 0;
	return 1;
}
int main() {
	int m, n, k;
	scanf("%d%d%d", &m, &n, &k);
	int popOrder[1000];
	for(int i = 0; i < k; i++) {
		for(int j = 0; j < n; j++)
			scanf("%d", &popOrder[j]);
		if(IsPopSeq(popOrder, m, n))
			printf("YES\n");
		else
			printf("NO\n");
	}

	return 0;
}