刷题总结——郁闷的出纳员（bzoj1503）

题目：

题目背景

NOI2004 DAY1 T1

题目描述

OIER 公司是一家大型专业化软件公司，有着数以万计的员工。作为一名出纳员，我的任务之一便是统计每位员工的工资。这本来是一份不错的工作，但是令人郁闷的是，我们的老板反复无常，经常调整员工的工资。如果他心情好，就可能把每位员工的工资加上一个相同的量。反之，如果心情不好，就可能把他们的工资扣除一个相同的量。我真不知道除了调工资他还做什么其它事情。

工资的频繁调整很让员工反感，尤其是集体扣除工资的时候，一旦某位员工发现自己的工资已经低于了合同规定的工资下界，他就会立刻气愤地离开公司，并且再也不会回来了。每位员工的工资下界都是统一规定的。每当一个人离开公司，我就要从电脑中把他的工资档案删去，同样，每当公司招聘了一位新员工，我就得为他新建一个工资档案。

老板经常到我这边来询问工资情况，他并不问具体某位员工的工资情况，而是问现在工资第 k 多的员工拿多少工资。每当这时，我就不得不对数万个员工进行一次漫长的排序，然后告诉他答案。

好了，现在你已经对我的工作了解不少了。正如你猜的那样，我想请你编一个工资统计程序。怎么样，不是很困难吧？

输入格式

第一行有两个非负整数 n 和 min 。n 表示下面有多少条命令，min 表示工资下界。
接下来的 n 行，每行表示一条命令。命令可以是以下四种之一：

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" 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“下划线（ _ ）”表示一个空格，I 命令、A 命令、S 命令中的 k 是一个非负整数，F 命令中的 k 是一个正整数。

在初始时，可以认为公司里一个员工也没有。

输出格式

输出的行数为 F 命令的条数加一。

对于每条 F 命令，你的程序要输出一行，仅包含一个整数，为当前工资第 k 多的员工所拿的工资数，如果 k 大于目前员工的数目，则输出 -1 。

输出的最后一行包含一个整数，为离开公司的员工的总数。

样例数据 1

输入　 [复制]

 

9 10 
I 60 
I 70 
S 50 
F 2 
I 30 
S 15 
A 5 
F 1 
F 2

输出

10 
20 
-1 
2

备注

【数据范围】
I 命令的条数不超过 100000
A 命令和 S 命令的总条数不超过 100
F 命令的条数不超过 100000
每次工资调整的调整量不超过 1000
新员工的工资不超过 100000

【评分方法】
对于每个测试点，如果你输出文件的行数不正确，或者输出文件中含有非法字符，得分为 0 。

否则你的得分按如下方法计算：

• 如果对于所有的 F 命令，你都输出了正确的答案，并且最后输出的离开公司的人数也是正确的，你将得到 10 分；
• 如果你只对所有的 F 命令输出了正确答案，得 6 分；
• 如果只有离开公司的人数是正确的，得 4 分；否则得 0 分。

题解：

重要的事情先说下：

仔细审题！！！！！

做的时候以为加入公司就走的员工是要算入总数的·····结果tm不算·····感觉题目也没说清楚啊···

其余的就是splay的基本操作了···另外这道题有个很巧妙的地方是记录一个sum维护修改公司的总和，并根据sum不是直接修改所有员工的工资数，而是修改加入的人的工资数和每次判断离开公司的人时修改minn（具体看代码）····，这样就能节约复杂度了···

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int N=3e5+;
int val[N],father[N],cnt[N],son[N][],size[N],root,tot;
int n,a,minn,totl=,sum=;
long long totcut=;
char s[];
inline int R()
{
  char c;int f=;
  for(c=getchar();c<''||c>'';c=getchar());
  for(;c<=''&&c>='';c=getchar())
    f=(f<<)+(f<<)+c-'';
  return f;
}
inline void clear(int now)
{
  size[now]=son[now][]=son[now][]=val[now]=cnt[now]=father[now]=;
}
inline void update(int now)
{
  if(now)
  {
    size[now]=cnt[now];
    if(son[now][])  size[now]+=size[son[now][]];
    if(son[now][])  size[now]+=size[son[now][]];
  }
}
inline int get(int a)
{
  return son[father[a]][]==a;
}
inline void rotate(int now)
{
  int fa=father[now],ofa=father[fa],which=get(now);
  son[fa][which]=son[now][which^],father[son[fa][which]]=fa;
  son[now][which^]=fa,father[fa]=now,father[now]=ofa;
  if(ofa)  son[ofa][son[ofa][]==fa]=now;
  update(fa),update(now);
}
inline void splay(int now)
{
  while(father[now])
  {
    if(father[father[now]])  rotate(get(now)==get(father[now])?father[now]:now);
    rotate(now);
  }
  root=now;
}
inline void insert(int x)
{
  int now=root,last=;
  while(true)
  {
    if(!now)
    {
      now=++tot;size[now]=cnt[now]=;father[now]=last;val[now]=x;
      son[last][val[now]>val[last]]=now;update(last);splay(now);
      break;
    }
    if(val[now]==x)
    {
      cnt[now]++;update(now);update(last);splay(now);
      break;
    }
    last=now;now=son[now][x>val[now]];
  }
}
inline int findx(int x)
{
  int now=root;
  while(true)
  {
    if(x<=size[son[now][]])  now=son[now][];
    else
    {
      int temp=size[son[now][]]+cnt[now];
      if(x<=temp)  return val[now];
      x-=temp;now=son[now][];
    }
  }
}
inline int Delete()
{
  int temp=size[son[root][]]+cnt[root];int oldroot=root;
  root=son[oldroot][];father[root]=;
  clear(oldroot);
  return temp;
}
int main()
{
  //freopen("a.in","r",stdin);
  n=R(),minn=R();
  while(n--)
  {
    scanf("%s%d",s,&a);
    if(s[]=='I')
    {
      if(a-sum>(minn-sum-))
        insert(a-sum),totl++;
    }
    if(s[]=='A')
      sum+=a;
    if(s[]=='S')
      sum-=a;
    if(s[]=='F')
    {
      if(totl<a)  cout<<"-1"<<endl;
      else
      {
        int ans=findx(totl-a+);
        cout<<ans+sum<<endl;
      }
    }
    int limit=minn-sum-;insert(limit);
    int temp=Delete();
    totcut+=(long long)(temp-);
    totl=totl-temp+;
  }
  cout<<totcut<<endl;
  return ;
}