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02-线性结构4 Pop Sequence

题目描述

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
结尾无空行
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Sample Output:

YES
NO
NO
YES
NO
结尾无空行
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代码实现

#include <stdio.h>
#include <stdbool.h>

typedef struct {
    int data[1000];
    int count;
} MyStack;

MyStack stack;

// 建立栈
void CreateStack()
{
    stack.count = 0;
}

// Push
void PushStack(int num)
{
    stack.data[stack.count] = num;
    stack.count++;
}

// Pop
void PopStack()
{
    stack.count--;
}

bool CheckPopInfo(int *num, int n, int m)
{
    int cur = 1;
    for (int i = 0; i < n; i++) {
        // stack为空或者栈顶元素与num[i]不相等,则继续入栈
        while (stack.count == 0 || stack.data[stack.count-1] != num[i]) {
            if (cur > n || stack.count >= m) {
                break;   // 过限
            }
            stack.data[stack.count] = cur++;
            stack.count++;
        }

        // 判断是否符合,不等于num[i]或者stack是空说明是break跳出来的,不符合
        if (stack.data[stack.count-1] != num[i] || stack.count == 0) {
            return false;
        }

        // 继续查看下一个数
        stack.count--;   // 出栈
    }
    return true;
}

int main()
{
    int m;
    int n;
    int k;
    scanf("%d%d%d", &m, &n, &k);
    for (int j = 0; j < k; j++) {
        int num[1000] = {0};
        for (int i = 0; i < n; i++) {
            scanf("%d", &num[i]);
        }

        CreateStack();
        if (CheckPopInfo(num, n, m)) {
            printf("YES\n");
        } else {
            printf("NO\n");
        }
    }
    
    return 0;
}
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