8VC Venture Cup 2016 - Elimination Round G. Raffles 线段树

G. Raffles

题目连接：

http://www.codeforces.com/contest/626/problem/G

Description

Johnny is at a carnival which has n raffles. Raffle i has a prize with value pi. Each participant can put tickets in whichever raffles they choose (they may have more than one ticket in a single raffle). At the end of the carnival, one ticket is selected at random from each raffle, and the owner of the ticket wins the associated prize. A single person can win multiple prizes from different raffles.

However, county rules prevent any one participant from owning more than half the tickets in a single raffle, i.e. putting more tickets in the raffle than all the other participants combined. To help combat this (and possibly win some prizes), the organizers started by placing a single ticket in each raffle, which they will never remove.

Johnny bought t tickets and is wondering where to place them. Currently, there are a total of li tickets in the i-th raffle. He watches as other participants place tickets and modify their decisions and, at every moment in time, wants to know how much he can possibly earn. Find the maximum possible expected value of Johnny's winnings at each moment if he distributes his tickets optimally. Johnny may redistribute all of his tickets arbitrarily between each update, but he may not place more than t tickets total or have more tickets in a single raffle than all other participants combined.

Input

The first line contains two integers n, t, and q (1 ≤ n, t, q ≤ 200 000) — the number of raffles, the number of tickets Johnny has, and the total number of updates, respectively.

The second line contains n space-separated integers pi (1 ≤ pi ≤ 1000) — the value of the i-th prize.

The third line contains n space-separated integers li (1 ≤ li ≤ 1000) — the number of tickets initially in the i-th raffle.

The last q lines contain the descriptions of the updates. Each description contains two integers tk, rk (1 ≤ tk ≤ 2, 1 ≤ rk ≤ n) — the type of the update and the raffle number. An update of type 1 represents another participant adding a ticket to raffle rk. An update of type 2 represents another participant removing a ticket from raffle rk.

It is guaranteed that, after each update, each raffle has at least 1 ticket (not including Johnny's) in it.

Output

Print q lines, each containing a single real number — the maximum expected value of Johnny's winnings after the k-th update. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Sample Input

2 1 3
4 5
1 2
1 1
1 2
2 1

Sample Output

1.666666667
1.333333333
2.000000000

Hint

题意

有n个奖池，每个奖池价值p[i]，你有t张票

现在每个奖池里面已经有了y[i]张票，然后你可以向每个奖池里面投入x[i]张票，当然x[i]的和需要小于等于t

然后你可以获得p[i]*x[i]/(y[i]+x[i])元钱（其实题意是你有x[i]/(y[i]+x[i])的概率获得p[i]元），但是有规定，你最多获得p[i]/2元

问你怎么投资，可以获得最多的钱

有q次修改

修改有两个操作:
1.使得y[i]+1
2.使得y[i]-1

每次修改完后，输出答案。

题解：

简单思考一下，假设没有修改操作的话，我们可以用一个堆来维护，每一块钱投入进当前我能够赚取最多的奖池就好了。

修改操作也是一样的，我修改之后，我讨论一下我当前的状态，只要我的一块钱从一个地方移动到另外一个地方，能够赚钱的话，我就去移动就好了

现在我就用线段树去维护，我从一个地方失去一块钱，我亏多少，从一个地方投入一块钱，我赚多少

然后从亏最少的地方转移到最多的地方就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+6;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int x[maxn],y[maxn],p[maxn];
struct treenode
{
  int L , R  ;
  double Up,Down,Max,Min,ans;
  void updata()
  {
        ans=1.0*p[L]*min(1.0*x[L]/(x[L]+y[L]),0.5);
      if(x[L]>=y[L])Up=0;
      else
      {
          Up=1.0*p[L]*(x[L]+1.0)/(x[L]+y[L]+1.0);
          Up-=1.0*p[L]*x[L]/(x[L]+y[L]);
      }
      if(x[L])
      {
          if(x[L]>y[L])Down=0;
          else
          {
              Down=1.0*p[L]*x[L]/(x[L]+y[L]);
              Down-=1.0*p[L]*(x[L]-1.0)/(x[L]-1.0+y[L]);
          }
      }
      else
        Down=1e18;
  }
};

treenode tree[maxn*4];
inline void push_up(int o)
{
    tree[o].ans=tree[o<<1].ans+tree[o<<1|1].ans;
    tree[o].Up=max(tree[o<<1].Up,tree[o<<1|1].Up);
    tree[o].Down=min(tree[o<<1].Down,tree[o<<1|1].Down);
    if(tree[o<<1].Up>tree[o<<1|1].Up)
        tree[o].Max=tree[o<<1].Max;
    else
        tree[o].Max=tree[o<<1|1].Max;
    if(tree[o<<1].Down<tree[o<<1|1].Down)
        tree[o].Min=tree[o<<1].Min;
    else
        tree[o].Min=tree[o<<1|1].Min;
}

inline void build_tree(int L , int R , int o)
{
	tree[o].L = L , tree[o].R = R, tree[o].ans=0;
	if(L==R)
        tree[o].Min=tree[o].Max=L,tree[o].updata();
	if (R > L)
	{
		int mid = (L+R) >> 1;
		build_tree(L,mid,o*2);
		build_tree(mid+1,R,o*2+1);
		push_up(o);
	}
}

inline void updata(int QL,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (L==R)
    {

        tree[o].updata();
    }
	else
	{
		int mid = (L+R)>>1;
		if (QL <= mid) updata(QL,o*2);
		else updata(QL,o*2+1);
		push_up(o);
	}
}

int main()
{
    int n,t,q,mx,mi;
    scanf("%d%d%d",&n,&t,&q);
    for(int i=1;i<=n;i++)
        p[i]=read();
    for(int i=1;i<=n;i++)
        y[i]=read();
    build_tree(1,n,1);
    while(t--)mx=tree[1].Max,x[mx]++,updata(mx,1);
    while(q--)
    {
        int type,r;type=read(),r=read();
        if(type==1)y[r]++;else y[r]--;
        updata(r,1);
        while(1)
        {
            int mx = tree[1].Max;
            int mi = tree[1].Min;
            //cout<<mx<<" "<<mi<<" "<<tree[1].Up<<" "<<tree[1].Down<<endl;
            if(tree[1].Up<=tree[1].Down)break;
            x[mx]++,x[mi]--;
            updata(mx,1);
            updata(mi,1);
        }
        printf("%.12f\n",tree[1].ans);
    }
}