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
本题大意：给出若干组数组，检测每组数组（每个数字池）能否通过指定大小的栈pop得到，入栈顺序：1到N。
输入第一行：M（栈的最大容量）N（要检测的数字有多少个）K（有多少组要检测的数字）；后K行每行N个数字，代表要检测的数组；输出：如果可以通过栈的pop得到则输出YES，否则输出NO
如： 5 6 4 3 7 2 1：push:1,2,3,4,5; pop:5, push:6, pop:6, pop:4, pop:3, push:7, pop:7, pop:2, pop:1，即可得到。
解题思路：设置两个指针分别指向数字池的首端和栈顶，根据先入后出原则，用数组模拟栈，栈顶应该是数组的最后一个元素。如果栈顶元素小于当前数字池的那个元素，则向栈里压入一个新数字，如果栈顶元素等于当前数字池的那个数字，则出栈一个数组，同时数字池指针向后移一位，表示下一个检测的元素，如果栈顶元素大于数字池指针指向的元素，则说明数字池的元素被放在了栈顶下面，无法弹出，则为NO；注意栈满的情况，在入栈之前应该先判断栈是否满了。
c语言版：
#include <stdio.h>
#include <string.h>
int main() {
int M, N, K;
scanf("%d %d %d", &M, &N, &K);
int stack[M], pool[N];
while(K--) {
//初始化要检测的数字池
int flag = 0;
for (int i = 0; i < N; ++i) {
scanf("%d", &pool[i]);
}
int pPool = 0, pStack = -1, num = 1;
while (pPool < N) {
if (pStack == -1) {
// 栈空
stack[++pStack] = num++;
} else {
// 栈不为空，检查栈顶元素和数字池顶元素是否匹配
if (stack[pStack] == pool[pPool]) {
// 匹配
pStack--;
pPool++;
} else if (stack[pStack] < pool[pPool]) {
// 栈顶元素比较小，则先判断后入栈
if ((++pStack) == M) {
flag = 1;
break;
}
stack[pStack] = num++;
} else {
flag = 1;
break;
}
}
}
if (flag) printf("NO\n");
else printf("YES\n");
}
return 0;
}
/*
5 7 1
1 7 6 5 4 3 2
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
*/
c++版
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main() {
int M, N, K;
cin >> M >> N >> K;
while(K--) {
vector<int> pool(N, 0);
// 输入一个数字池
for (int i = 0; i < N; ++i) {
cin >> pool[i];
}
stack<int> stack1;
int pPool = 0, num = 1, flag = 0;
while (pPool < N) {
if (stack1.empty()) {
// 判断是不是为空，如果是，则入栈
stack1.push(num++);
} else {
// 不为空，看下栈顶元素和数字池的当前元素是不是一样
if (stack1.top() == pool[pPool]) {
// 一样则出栈一个元素，数字池的指针指向下一个元素
stack1.pop();
pPool++;
} else if (stack1.top() < pool[pPool]) {
// 栈顶元素比较小，则先检测是不是满栈，再入栈
if (stack1.size() == M) {
flag = 1;
break;
} else {
stack1.push(num++);
}
} else {
// 栈顶元素比较大则不能产生数字池的数组的顺序
flag = 1;
break;
}
}
}
if (flag) cout << "NO" << endl;
else cout << "YES" << endl;
}
return 0;
}
/*
5 7 5
1 7 6 5 4 3 2
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
*/