PAT 1051
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
一,以前解法
#include<cstdio>
#include<stack>
using namespace std;
//到14:45
stack<int> st;
int arr[1010];
int main() {
int m = 0;
int n = 0;
int k = 0;
scanf("%d %d %d", &m, &n, &k);
while (k--) {
while (!st.empty()) {
st.pop();
}
int count = 0;
int flag = 0;
int max = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
for (int i = 0; i < n; i++) {
if (!st.empty()&&arr[i] > st.top()) {
for (int j = max + 1; j <= arr[i]; j++) {
st.push(j);
if (arr[i] > max) {
max = arr[i];
}
if (count > 5) {
flag = 1;
break;
}
}
}
if (st.empty()) {
for (int j = max+1; j <= arr[i]; j++) {
st.push(j);
count++;
if (arr[i] > max) {
max = arr[i];
}
if (count > 5) {
flag = 1;
break;
}
}
}
if (arr[i] == st.top()) {
st.pop();
count--;
}
else {
flag = 1;
break;
}
}
if (flag == 0&&st.empty()) {
printf("YES\n");
}
else {
printf("NO\n");
}
}
return 0;
}
二,如今解法
#include<cstdio>
#include<stack>
using namespace std;
//到14:45
stack<int> st;
int arr[1010];
int main() {
int m = 0;
int n = 0;
int k = 0;
int flag = 1;
scanf("%d %d %d", &m, &n, &k);
for (int i = 0; i < k; i++) {
while (!st.empty()) {
st.pop();
}
for (int j = 0; j < n; j++) {
scanf("%d", &arr[j]);
}
int current = 0;
for (int i = 1; i <= n; i++) {
st.push(i);
if (st.size() > m) {
break;
flag = 0;
}
while (!st.empty() && st.top() == arr[current]) {
st.pop();
current++;
}
}
if (flag == 1 && st.empty()) {
printf("YES\n");
}
else {
printf("NO\n");
}
}
return 0;
}
三,总结
以前解法是对要检验的数组的每个值进行遍历,然后对栈进行相应的处理,有不合格的直接退出。栈的增加要以目前的数组扫描到的为界限,会引发一系列问题,例如栈提前空了怎么办,当前扫描的值与栈顶元素大小的比较等。
正确解法是以顺序插入为驱动,与当前扫描的值一致的时候pop,然后自动扫描下一个,不一致继续顺序插入,没有前一种解法的冗余。
四,Note
要在pop()和top()操作前判空。