1051 Pop Sequence(25 分)
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
解题思路:
模拟栈的输入,判断时候是合法的输出序列。
先把输入的序列接收进数组vi。然后按顺序1~n把数字进栈,每进入一个数字,判断有没有超过最大范围,超过了就break。如果没超过,设cur = 1,从数组的第一个数字开始,看看是否与栈顶元素相等,while相等就一直弹出栈,不相等就继续按顺序把数字压入栈~~~最后根据变量flag的bool值输出yes或者no。
#include<cstdio>
#include<stack>
#include<vector>
using namespace std;
int main(){
int n,m,k;
scanf("%d %d %d",&m,&n,&k);
for(int i=0;i<k;i++) {
bool flag=false;
stack<int>st;
vector<int>vi(n+1);
for(int j=1;j<=n;j++){
scanf("%d",&vi[j]);
}
int cur=1;
for(int j=1;j<=n;j++){
st.push(j) ;
if(st.size() >m) break;
while(!st.empty()&&st.top() ==vi[cur] ){
st.pop() ;
cur++;
}
}
if(cur==n+1) flag=true;
if(flag==true) printf("YES\n");
else printf("NO\n");
}
return 0;
}
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