Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., Nand 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.
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.
For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.
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
YES
NO
NO
YES
NO
题⽬⼤意:有个容量限制为m的栈,分别把1,2,3,…,n⼊栈,
给出⼀个系列出栈顺序,问这些出栈 顺序是否可能~
分析:按照要求进⾏模拟。先把输⼊的序列接收进数组v。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current = 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
#include
#include
using namespace std;
int main() {
int m, n, k;
scanf("%d %d %d", &m, &n, &k);
for( int i = 0; i < k; i++) {
bool flag = false;
stack s;
vector v(n + 1);
for( int j = 1; j <= n; j++)
scanf("%d", &v[j]);
int current = 1;
for( int j = 1; j <= n; j++) {
s.push(j);
if( s.size() > m ) break;
while( !s.empty() && s.top() == v[current] ) {
s.pop();
current++;
}
}
if( current == n + 1 )
flag = true;
if( flag )
printf("YES\n");
else printf("NO\n");
}
return 0;
}