并不对劲的bzoj1758:p4292:[WC2010]重建计划

题目大意

\(n\)(\(n\leq10^5\))个点的一棵树,有边权\(w\),给定\(l,r\),求边数在\([l,r]\)中的路径的平均边权的最大值

题解

二分答案,判断时将边权变成\(w-mid\),判断是否存在一条边权总和非负且边数在[l,r]的路径
\(f(i,j)\)表示从点\(i\)往下走\(j\)条边的边权总和最多是多少
则有\(f(i,j)=max_{v\in son(i)}\{f(v,j-1)+w(i,v)\}\)
\(ans(i)=max_{v\in son(i) , l\leq k+j+1\leq r}\{f(v,k)+f(i,j)+w(i,v)\}\)
考虑长链剖分,将\(f(i,j)\)存到\(F(i,j)=dfn[i]+j\)的位置
需要注意的是,算完点\(i\)的长儿子后,需要将\(F(i,1)\)\(F(i,dep[i])\)的值+\(w(i,v)\)
看上去常数比点分治小,事实上。。。

代码
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define view(u,k) for(int k=fir[u];~k;k=nxt[k])
#define maxn 100010 
#define maxm (maxn<<1)
#define F(x,y) (dfn[x]+y)
#define inf (1e11)
#define ls (u<<1)
#define rs (u<<1|1)
#define mi (L+R>>1)
#define eps (1e-4)
using namespace std;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)&&ch!='-')ch=getchar();
    if(ch=='-')f=-1,ch=getchar();
    while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return x*f;
}
void write(int x)
{
    if(x==0){putchar('0'),putchar('\n');return;}
    int f=0;char ch[20];
    if(x<0)putchar('-'),x=-x;
    while(x)ch[++f]=x%10+'0',x/=10;
    while(f)putchar(ch[f--]);
    putchar('\n');
    return;
}
int n,l,r,cnt,fir[maxn],nxt[maxm],v[maxm],son[maxn],fa[maxn],dfn[maxn],tim,dep[maxn];
double toso[maxn],w[maxm],tr[maxn<<2],mk[maxn<<2],ans,fakeans,mid;
inline void ade(int u1,int v1,int w1){w[cnt]=(double)w1,v[cnt]=v1,nxt[cnt]=fir[u1],fir[u1]=cnt++;}
void getdep(int u)
{
    view(u,k)if(v[k]!=fa[u])
    {
        fa[v[k]]=u,getdep(v[k]),dep[u]=max(dep[v[k]]+1,dep[u]);
        if(!son[u]||dep[v[k]]>dep[son[u]])son[u]=v[k],toso[u]=w[k];
    }
}
void getdfn(int u)
{
    dfn[u]=++tim;
    if(son[u])getdfn(son[u]);
    view(u,k)if(v[k]!=fa[u]&&v[k]!=son[u])getdfn(v[k]);
}
inline void mark(int u,double k){if(tr[u]!=-inf)mk[u]+=k,tr[u]+=k;}
inline void pd(int u){if(mk[u]==-inf){tr[ls]=tr[rs]=-inf,mk[ls]=mk[rs]=-inf,mk[u]=0;}if(mk[u]){mark(ls,mk[u]),mark(rs,mk[u]),mk[u]=0;}}
void add(int u,int L,int R,int x,int y,double k)
{
    if(y<L||R<x||y<x)return ;
    if(x<=L&&R<=y)return mark(u,k);
    pd(u);
    if(x<=mi)add(ls,L,mi,x,y,k);
    if(y>mi)add(rs,mi+1,R,x,y,k);
    tr[u]=max(tr[ls],tr[rs]);
}
void chg(int u,int L,int R,int x,double k)
{
    if(x<L||R<x)return;
    if(x<=L&&R<=x){tr[u]=max(tr[u],k);return;}
    pd(u);
    if(x<=mi)chg(ls,L,mi,x,k);
    if(x>mi)chg(rs,mi+1,R,x,k);
    tr[u]=max(tr[ls],tr[rs]);
}
double ask(int u,int L,int R,int x,int y)
{
    if(y<L||R<x||y<x)return -inf;
    if(x<=L&&R<=y)return tr[u];
    pd(u);
    double res=-inf;
    if(x<=mi)res=ask(ls,L,mi,x,y);
    if(y>mi)res=max(res,ask(rs,mi+1,R,x,y));
    return res;
}
double tmp[maxn];
void getans(int u)
{
    if(son[u])getans(son[u]),add(1,1,n,F(u,1),F(u,dep[u]),toso[u]-mid);
    if(fakeans>=0)return;
    chg(1,1,n,F(u,0),0);
    int Llim=max(0,l),Rlim=min(dep[u],r);
    if(Llim<=Rlim)
    {
        double tmp=ask(1,1,n,F(u,Llim),F(u,Rlim));
        fakeans=max(fakeans,tmp);
        if(fakeans>=0)return;
    }
    view(u,k)if(v[k]!=fa[u]&&v[k]!=son[u])
    {
        getans(v[k]);
        if(fakeans>=0)return;
        rep(i,0,dep[v[k]])
        {
            Llim=max(0,l-i-1),Rlim=min(r-i-1,dep[u]);
            tmp[i]=ask(1,1,n,F(v[k],i),F(v[k],i))+w[k]-mid;
            if(Llim<=Rlim)
            {	
                double tmp2=ask(1,1,n,F(u,Llim),F(u,Rlim));
                fakeans=max(fakeans,tmp2+tmp[i]);
                if(fakeans>=0)return;
            }
        }
        rep(i,0,dep[v[k]])chg(1,1,n,F(u,i+1),tmp[i]);
    }
}
int main()
{
    memset(fir,-1,sizeof(fir));
    n=read(),l=read(),r=read();double L=inf,R=0;
    rep(i,1,n-1){int x=read(),y=read(),z=read();ade(x,y,z),ade(y,x,z),L=min(L,(double)z),R=max(R,(double)z);}
    getdep(1),getdfn(1);
    while(R-L>=eps)
    {
        mid=(R+L)/2.0;
        fakeans=-inf;tr[1]=mk[1]=-inf;
        getans(1);
        if(fakeans>=0)ans=max(ans,mid),L=mid+eps;
        else R=mid-eps;
    }
    printf("%.3lf",ans);
    return 0;
}
/*
4 
2 3 
1 2 1 
1 3 2 
1 4 3 
*/
posted @ 2019-03-07 16:37  echo6342  阅读(131)  评论(0编辑  收藏  举报