pat02-线性结构4. Pop Sequence (25)

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

---

提交代码

#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
using namespace std;
int line[1005];
int main(){
    //freopen("D:\input.txt","r",stdin);
    int m,n,k;
    int i,j,num,hav,count;
    scanf("%d %d %d",&m,&n,&k);
    while(k--){
        stack<int> s;
        count=0;
        hav=0;
        for(i=0;i<n;i++)  
            scanf("%d",&line[i]);
        for(i=0;i<n;i++){
            num=line[i];            
            //hav表示的是入栈过的最大数 
            //当前栈顶元素<=hav，凡是<=hav的元素都已经进入到栈中 
            //1.num>hav时，num肯定>当前的栈顶元素，故hav+1到num的元素要进栈&&栈的最大元素个数<=m 
            //2.不满足1时，num<hav&&num<=当前栈顶元素，不断退栈，直到退到满足条件的元素或者栈空 
            //3.
            if(hav<num&&num-hav+count<=m){//num>hav时，num肯定>当前的栈顶元素，故hav+1到num的元素要进栈&&栈的最大元素个数<=m 
                for(j=hav+1;j<num;j++){
                    
                    //cout<<"j:  "<<j<<endl;
                    
                    s.push(j);
                    count++; 
                }
                hav=num; 
            }
            else  if(!s.empty()&&s.top()>=num){
                //不满足1时，num<hav&&num<=当前栈顶元素，不断退栈，直到退到满足条件的元素或者栈空
                       while(!s.empty()&&s.top()!=num){
                                s.pop();
                                count--;
                        }
                       if(s.empty()){//栈空说明这个元素之前已经pop过，不符合题意 
                                break;
                       }
                       else{//找到满足的元素 
                                s.pop();
                                count--;
                       }
                }else{//其他情况不符合题意 
                    break;
                }
        }
        if(i==n){
            printf("YES
");
        } 
        else{
            printf("NO
");
        }
    }
    return 0;
}