coderforces round#420-C Okabe and Boxes

最新推荐文章于 2023-05-09 14:27:36 发布

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解决一个涉及盒子堆叠与移除的问题，通过最少次重新排序确保盒子按指定顺序被移除。

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.

Input

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.

Output

Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.

Examples

input
3
add 1
remove
add 2
add 3
remove
remove

output
1

input
7
add 3
add 2
add 1
remove
add 4
remove
remove
remove
add 6
add 7
add 5
remove
remove
remove

output
2

题意是：有1-n标号的箱子和2*n个操作以及一个栈，让你用最少的排序次数将箱子出栈的顺序为1到N，这里排序是指，当出现出栈的数字不对时，就应该将栈里的元素排序一次，然后取出合法的元素，由于题目保证输入合法，所以排序之后只管出栈就好。

思路：考虑到应该要在出栈的时候操作，所以我们只要建一个栈模拟一下整个过程就好，由于输入合法，如果要出栈的数恰好是我们要的数，直接栈顶元素删除，如果不是我们就把栈清空，（相当于将栈里元素从大到小排好序），如果下一次进来，栈为空，说明栈里元素是合法的有序队列，直接弹出，（题目保证输入合法）

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define siz 300005
using namespace std;
int n;
int sta[siz],top,ans,data;
void operator1(){
    int x;
    scanf("%d",&x);
    sta[++top]=x;

}
void operator2(){
    if(top==0){ ++data;return ;}
    else{
        if(sta[top]==data){
            --top;
            ++data;
        }
        else{
            ++ans;
            ++data;
            top=0;
        }
    }
}
int main()
{
    char s[15];
    while(~scanf("%d",&n))
    {
        top=0,ans=0;
        data=1;
       int t=2*n;
       while(t--){
            scanf("%s",s);
            if(s[0]=='a'){
                operator1();
            }
            else{
                operator2();
            }
       }
       cout<<ans<<endl;
    }
    return 0;
}