Codeforces Round #420 (Div. 2) C. Okabe and Boxes 栈+last标记+贪心

C. Okabe and Boxes

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

3add 1removeadd 2add 3removeremove

output

1

input

7add 3add 2add 1removeadd 4removeremoveremoveadd 6add 7add 5removeremoveremove

output

2

Note

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.

Source

Codeforces Round #420 (Div. 2)

My Solution

题意：有1~n这n个数，给出一个stack的push和pop的序列，要求在执行的过程中用尽可能少的重排次数，使得能够使pop的顺序是1~n的顺序。

栈+last标记+贪心

1~2*n个操作，用一个last标记最近一次重排的位置，初始为last = 0。

然后用一个栈表示上一次重排后push如的元素的栈结构。

每次remove，如果当前需要pop的cnt == top则直接从stack pop掉，

否则，如果栈空，且mp[cnt]出现的位置在last之前，则可以直接pop，

           不然则需要一次重排，然后pop，并且清空当前栈。

复杂度 O(n)

#include <iostream>#include <cstdio>#include <string>#include <stack>#include <cstring>using namespace std;typedef long long LL;const int MAXN = 3e5 + 8;int mp[MAXN];string s;stack<int> sta;int main(){    #ifdef LOCAL    freopen("c.txt", "r", stdin);    //freopen("c.out", "w", stdout);    int T = 4;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int n, i, j, val, top = -1, last = 0, ans = 0, cnt = 1;    cin >> n;    for(i = 1; i <= 2*n; i++){        cin >> s;        if(s[0] == 'a'){            cin >> val;            mp[val] = i;            top = val;            sta.push(val);        }        else{            if(cnt != top){                if(mp[cnt] <= last && sta.empty()){                    ;                }                else{                    last = i;                    ans++;                }                while(!sta.empty()) sta.pop();                top = -1;            }            else{                sta.pop();                if(!sta.empty())top = sta.top();                else top = -1;            }            cnt++;        }    }    cout << ans << endl;    #ifdef LOCAL    while(!sta.empty()) sta.pop();    memset(mp, 0, sizeof mp);    cout << endl;    }    #endif // LOCAL    return 0;}

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