BZOJ1758: [Wc2010]重建计划（01分数规划+点分治+单调队列）

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1758

01分数规划，所以我们对每个重心进行二分。于是问题转化为Σw[e]-mid>=0， 对于一棵子树维护点的dep，dis，并用队列q存下来。令mx[i]表示当前dep为i的最大权值，维护一个单调队列dq，维护当前符合条件的mx，当我们从q的队尾向前扫时，它的dep是递减的，利用这个性质维护单调队列，最后更新一遍mx。具体看代码吧TAT

代码：

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<queue>
#include<set>
#include<cmath>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 209000
#define inf int(1e9)
#define mm 1000000007
#define esp 1e-6
using namespace std;
#define ll long long
struct data{int obj,pre; double c;
}e[maxn*2];
int head[maxn],s[maxn],q[maxn],dq[maxn],dep[maxn],fa[maxn],vis[maxn];
double ans,lim,dis[maxn],mx[maxn];
int n,m,tot,sum,mn,rt,L,U;
int read(){
    int x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
    return x*f;
}
void insert(int x,int y,double z){
    e[++tot].obj=y; e[tot].c=z; e[tot].pre=head[x]; head[x]=tot;
}
void dfs(int u,int fa){
    s[u]=1; int mx=0;
    for (int j=head[u];j;j=e[j].pre){
        int v=e[j].obj;
        if (v!=fa&&!vis[v]) {
            dfs(v,u);
            s[u]+=s[v];
            mx=max(mx,s[v]);
        }
    }
    mx=max(mx,sum-mx);
    if (mx<mn) mn=mx,rt=u;
}
bool go(int u,double mid){
    int up=0;
    for (int j=head[u];j;j=e[j].pre){
        int v=e[j].obj;  
        if (vis[v]) continue;
        dep[v]=1; dis[v]=e[j].c-mid; fa[v]=u;
        int l=0,r=1; q[1]=v;
        while (l<r){
            int now=q[++l];
            for (int k=head[now];k;k=e[k].pre){
                int v=e[k].obj;
                if (v!=fa[now]&&!vis[v]){
                    fa[v]=now; dep[v]=dep[now]+1; dis[v]=dis[now]+e[k].c-mid;
                    q[++r]=v;
                }
            }
        }
        int tail=r; l=1; r=0; int now=up;
        rep(i,1,tail){
            while (dep[q[i]]+now>=L&&now>=0){
                while (l<=r&&mx[now]>mx[dq[r]]) r--;
                dq[++r]=now;
                now--;
            }
            while (l<=r&&dep[q[i]]+dq[l]>U) l++;
            if (l<=r&&dis[q[i]]+mx[dq[l]]>=0) return 1;
        }
        rep(i,up+1,dep[q[tail]]) mx[i]=-inf;
        rep(i,1,tail) {
            int now=dep[q[i]];
            mx[now]=max(mx[now],dis[q[i]]);
        }
        up=max(up,dep[q[tail]]);
    }
    return 0;
}
void jud(int u){
    double l=ans,r=lim;
    while (r-l>=0.0001){
        double mid=(l+r)/2;
        if (go(u,mid)) l=mid;
        else r=mid;
    }
    ans=l;
}
void solve(int u){
    mn=inf;
    dfs(u,0);
    u=rt;
    jud(u);
    vis[u]=1;
     
    for (int j=head[u];j;j=e[j].pre){
        int v=e[j].obj;
        if (!vis[v]) {
            sum=s[v];
            if (sum>L) solve(v);
        }
    }
}
int main(){
    n=read();
    L=read(); U=read();
    int x,y; double z;
    rep(i,1,n-1){
        x=read(); y=read(); scanf("%lf",&z);
        insert(x,y,z);
        insert(y,x,z);
        lim=max(lim,z);
    }
    sum=n;
    solve(1);
    printf("%.3lf\n",ans);
    return 0;
}