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02-线性结构4 Pop Sequence

发表于 分类于 Valine：
本文字数： 1.8k 阅读时长 ≈ 2 分钟

​ 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:

1
2
3
4
5
6
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:

1
2
3
4
5
YES
NO
NO
YES
NO

分析：

  • 本解法主要思路为逆向放置
  • 即将测试用例中的数从最后一个开始放置
  • 如果能在一定容量的堆栈下输出N~1
  • 将放与取操作取反,即可满足题意

编码：

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#include<stdio.h>
//本解法主要思路为逆向放置
//即将测试用例中的数从最后一个开始放置
//如果能在一定容量的堆栈下输出N~1
//将放与取操作取反
//即可满足题意
int main_p3(){
    int M,N,K;
    int target;
    // 1.用数组表示堆栈存储元素,top为栈顶指针
    int top;
    int data[1001];
    // 2.读取堆栈容量M,顺序数字N,测试序列K
    scanf("%d %d %d",&M,&N,&K);
    int b[N];
    int n ;
    // 3.重复读取K组序列
    while(K--){
        target = N;
        top = -1;
        n = N;
        // 4.读取一组测试序列，存放在数组b中
        while(n){
            scanf("%d",&b[N-n]);
            n--;
        }
        n = N-1;
        while(n+1){
            //5.依次将数组b中的元素压入堆栈data中
            top++;
            if(top >= M){
                break;
            }
            data[top] = b[n];
            // 6.如果
            while(data[top] == target&&top >-1){
                target--;
                //printf("%d",top);
                top--;
                // printf("%d",top);
            }
            n--;
        }

        //printf("%d",top);
        //printf("%d",target);
        if(top == -1){
            printf("YES");
        }else{
            printf("NO");
        }
        if(K != 0){
            printf("\n");
        }
    }
    return 0;
}