Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): MM (the maximum capacity of the stack), NN (the length of push sequence), and KK (the number of pop sequences to be checked). Then KK lines follow, each contains a pop sequence of NN 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

時間限制：400ms
記憶體限制：64MB
程式碼長度限制：16kB
判題程式：系統預設
作者：陳越
單位：浙江大學
C語言順序棧實現

include

define MAXSIZE 1000

typedef struct{
int data[MAXSIZE];
int top;
}SqStack;

int InitStack(SqStack *s)
{
s->top = -1;
return 0;
}

int Push(SqStack *s, int e)
{
if(s->top==MAXSIZE)return 1;
s->top++;
s->data[s->top]=e;
return 0;
}

int Pop(SqStack *s){
int e;
if(s->top==-1)return 1;
e=s->data[s->top];
s->top–;
return e;
}

int main()
{
int m, n, k,i,h;
SqStack s,t;
scanf(“%d%d%d”,&m,&n,&k);
while(k–){
InitStack(&s);/初始化棧s和棧t/
InitStack(&t);
for(i=n-1;i>=0;i–){ /將給出的序列依次壓入棧t中(首元素為棧頂)/
scanf(“%d”,&(t.data[i]));
}
t.top=n-1;
i=0;
while(i<=n){ /模擬進棧出棧/
if(s.data[s.top]==t.data[t.top]&&s.top!=-1){/如果s的棧頂元素和t的棧頂元素相同/
Pop(&s);
Pop(&t);
}else if(s.top