Okabe and Super Hacker Daru are stacking and removing boxes. There are n boxes numbered from 1 to n. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2n commands: n of which are to add a box to the top of the stack, and n of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to n. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
The first line of input contains the integer n (1 ≤ n ≤ 3·105) — the number of boxes.
Each of the next 2n lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer x (1 ≤ x ≤ n) follows, indicating that Daru should add the box with number x to the top of the stack.
It is guaranteed that exactly n lines contain "add" operations, all the boxes added are distinct, and n lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
3 add 1 remove add 2 add 3 remove remove
1
7 add 3 add 2 add 1 remove add 4 remove remove remove add 6 add 7 add 5 remove remove remove
2
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack.
题意:模拟栈的操作,给定两种操作。
add:向栈中加入一个数。
remove:将栈顶的数弹出。
现在要求使栈中的元素按照1-n的顺序主次弹出,你可以在任意时刻将栈中的元素进行排序,问最少要进行多少次的排序。
分析:我们只需要在栈顶元素与所需弹出的元素不相同时进行一次排序,排序后可将栈内元素清空(因为已经符合了之后要弹出的顺序),记录排序次数即可。
代码如下:
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int MX = 3e5 + 5;
vector<int> vt;
int main(){
int n;
scanf("%d", &n);
int cnt = 1, ans = 0;
for(int i = 0; i < 2*n; i++){
char op[10];
scanf("%s", op);
if(op[0] == 'a'){
int x;
scanf("%d", &x);
vt.push_back(x);
}
else{
if(!vt.empty()){
if(vt.back() != cnt){
vt.clear();
ans++;
cnt++;
}
else{
vt.pop_back();
cnt++;
}
}
else cnt++;
}
}
printf("%d\n", ans);
return 0;
}