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[];
 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=;
         hav=;
         for(i=;i<n;i++)
             scanf("%d",&line[i]);
         for(i=;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+;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\n");
         }
         else{
             printf("NO\n");
         }
     }
     return ;
 }