线段树Codeforces Beta Round #99 (Div. 1)C

C. Mushroom Gnomes - 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Natalia was walking in the woods when she met a little mushroom gnome. The gnome told her the following story:

Everybody knows that the mushroom gnomes' power lies in the magic mushrooms that grow in the native woods of the gnomes. There aren trees and m magic mushrooms in the woods: thei-th tree grows at a point on a straight line with coordinatesa_i and has the height ofh_i, thej-th mushroom grows at the point with coordinatesb_j and has magical powersz_j.

But one day wild mushroommunchers, the sworn enemies of mushroom gnomes unleashed a terrible storm on their home forest. As a result, some of the trees began to fall and crush the magic mushrooms. The supreme oracle of mushroom gnomes calculated in advance the probability for each tree that it will fall to the left, to the right or will stand on. If the tree with the coordinatex and height h falls to the left, then all the mushrooms that belong to the right-open interval[x - h, x), are destroyed. If a tree falls to the right, then the mushrooms that belong to the left-open interval(x, x + h] are destroyed. Only those mushrooms that are not hit by a single tree survive.

Knowing that all the trees fall independently of each other (i.e., all the events are mutually independent, and besides, the trees do not interfere with other trees falling in an arbitrary direction), the supreme oracle was also able to quickly calculate what would be the expectation of the total power of the mushrooms which survived after the storm. His calculations ultimately saved the mushroom gnomes from imminent death.

Natalia, as a good Olympiad programmer, got interested in this story, and she decided to come up with a way to quickly calculate the expectation of the sum of the surviving mushrooms' power.

Input

The first line contains two integers n andm (1 ≤ n ≤ 10⁵,1 ≤ m ≤ 10⁴) — the number of trees and mushrooms, respectively.

Each of the next n lines contain four integers —a_i,h_i,l_i,r_i (|a_i| ≤ 10⁹,1 ≤ h_i ≤ 10⁹,0 ≤ l_i, r_i, l_i + r_i ≤ 100) which represent the coordinate of the i-th tree, its height, the percentage of the probabilities that the tree falls to the left and to the right, respectively (the remaining percentage is the probability that the tree will stand on).

Each of next m lines contain two integersb_j,z_j (|b_j| ≤ 10⁹,1 ≤ z_j ≤ 10³) which represent the coordinate and the magical power of thej-th mushroom, respectively.

An arbitrary number of trees and mushrooms can grow in one point.

Output

Print a real number — the expectation of the total magical power of the surviving mushrooms. The result is accepted with relative or absolute accuracy10^- 4.

Sample test(s)

Input

1 12 2 50 501 1

Output

0.5

Input

2 12 2 50 504 2 50 503 1

Output

0.25

Note

It is believed that the mushroom with the coordinate x belongs to the right-open interval[l, r) if and only if l ≤ x < r. Similarly, the mushroom with the coordinatex belongs to the left-open interval (l, r] if and only if l < x ≤ r.

In the first test the mushroom survives with the probability of 50%, depending on where the single tree falls.

In the second test the mushroom survives only if neither of the two trees falls on it. It occurs with the probability of 50% ×  50% = 25%.

Pretest №12 is the large test with 10⁵ trees and one mushroom.

思路：离散化然后区间更新

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=400010;int x[maxn],num,n,m,z[maxn/4],cnt,b[maxn/4];struct seg{    int l,r,c;    double p;    seg(){}    seg(int a,int b,int d,double pi):l(a),r(b),c(d),p(pi){}    bool operator < (const seg & a)const    {        if(l==a.l)return r<a.r;        return l<a.l;    }}s[maxn];struct IntervalTree{    double p[maxn<<2];    int setv[maxn<<2];    void build(int o,int l,int r)    {        p[o]=1.0;        setv[o]=0;        if(l==r)return;        int mid=(l+r)>>1;        build(o<<1,l,mid);        build(o<<1|1,mid+1,r);    }    void pushdown(int o,int l,int r)    {        if(l>=r)return;        if(setv[o])        {            p[o<<1]*=p[o];            p[o<<1|1]*=p[o];            p[o]=1;            setv[o<<1]=setv[o<<1|1]=1;            setv[o]=0;        }    }    void update(int o,int l,int r,int q1,int q2,double x)    {        if(q1<=l&&r<=q2)        {            p[o]*=x;            setv[o]=1;            return;        }        pushdown(o,l,r);        int mid=(l+r)>>1;        if(q1<=mid)update(o<<1,l,mid,q1,q2,x);        if(q2>mid)update(o<<1|1,mid+1,r,q1,q2,x);    }    double query(int o,int l,int r,int pos)    {        if(l==r)return p[o];        pushdown(o,l,r);        int mid=(l+r)>>1;        if(pos<=mid)return query(o<<1,l,mid,pos);        else return  query(o<<1|1,mid+1,r,pos);    }}tree;int main(){    freopen("in.txt","r",stdin);    scanf("%d%d",&n,&m);    num=cnt=0;    for(int i=1;i<=n;i++)    {        int a,h,l,r;        scanf("%d%d%d%d",&a,&h,&l,&r);        x[num++]=a-h;        x[num++]=a;        x[num++]=a+h;        s[cnt++]=seg(a-h,a,1,(100.0-l)/100.0);        s[cnt++]=seg(a,a+h,0,(100.0-r)/100.0);    }    for(int i=1;i<=m;i++)    {        scanf("%d%d",&b[i],&z[i]);        x[num++]=b[i];    }    sort(x,x+num);    int nx=unique(x,x+num)-x;    tree.build(1,0,nx);    for(int i=0;i<cnt;i++)    {        int l=lower_bound(x,x+nx,s[i].l)-x;        int r=lower_bound(x,x+nx,s[i].r)-x;        if(s[i].c)tree.update(1,0,nx,l,r-1,s[i].p);        else tree.update(1,0,nx,l+1,r,s[i].p);    }    double ans=0;    for(int i=1;i<=m;i++)    {        int pos=lower_bound(x,x+nx,b[i])-x;        ans+=tree.query(1,0,nx,pos)*z[i];    }    printf("%.4lf\n",ans);    return 0;}