02-线性结构4 Pop Sequence

02-线性结构4 Pop Sequence   (25分)

• 时间限制：400ms
• 内存限制：64MB
• 代码长度限制：16kB
• 判题程序：系统默认
• 作者：陈越
• 单位：浙江大学

https://pta.patest.cn/pta/test/3512/exam/4/question/62615

Given a stack which can keep MMM numbers at most. Push NNN numbers in the order of 1, 2, 3, ..., NNN 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 MMM is 5 and NNN 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): MMM (the maximum capacity of the stack), NNN (the length of push sequence), and KKK (the number of pop sequences to be checked). Then KKK lines follow, each contains a pop sequence of NNN 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

用回溯法（dfs）判断能否构造出目标序列

#include<bits/stdc++.h>
using namespace std;

int gode[1000];
int M,N,K,num;
stack<int> Sta;

bool dfs(int ipush,int ipop)//ipush,ipop为已完成的操作(push,pop)数
{
    if(ipop==N)//成功构造gode
        return true;
    if(ipush<N&&Sta.size()<M)
    {
        Sta.push(num++);
        if(dfs(ipush+1,ipop)) return true;
        num--;
        Sta.pop();
    }
    if(ipop<ipush&&!Sta.empty()&&Sta.top()==gode[ipop])//栈顶元素即为gode的下一个数时才执行pop
    {
        int t=Sta.top();
        Sta.pop();
        if(dfs(ipush,ipop+1)) return true;
        Sta.push(t);
    }
    return false;
}
int main()
{
    cin>>M>>N>>K;
    for(int i=0;i<K;i++)
    {
        num=1;
        for(int j=0;j<N;j++)
            cin>>gode[j];
        puts(dfs(0,0)?"YES":"NO");
    }
    return 0;
}