02-线性结构4 Pop Sequence （25 分)
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>
struct St {
int top;
int a[1050];
}stack;
int m, n, k;
void pop() {
stack.top--;
}
int stacktop() {
int temp = stack.a[stack.top];
return temp;
}
void push(int x) {
stack.top++;
stack.a[stack.top] = x;
}
int main() {
int t;
stack.top = -1;
scanf("%d %d %d", &m, &n, &k);
for (t = 0; t < k; t++) {
int i, j = 0, x = 1, flag = 0;
for (i = 0; i < n; i++) {
scanf("%d", &x);
if (flag == 1) continue;
while (j < x) {
if (stack.top + 1 >= m) {
flag = 1;
break;
}
j++;
push(j);
}
if (j == x) {
pop();
}
if (j > x) {
int y = stacktop();
if (y == x) {
pop();
}
else if (y != x) {
flag = 1;
}
}
}
if (flag == 1) printf("NO\n");
else if (flag == 0) printf("YES\n");
stack.top = -1;
}
return 0;
}