hdu 4729 树链剖分

思路:这个树链剖分其实还是比较明显的。将边按权值排序后插入线段树,然后用线段树查找区间中比某个数小的数和,以及这样的数的个数。当A<=B时,就全部建新的管子。

对于A>B的情况比较 建一条由S->T的管子后将这根管子容量扩到最大能得到的容量  与   将所有预算都用来扩大管子容量不建新管子得到的最大容量 做比较 ,选最大的。

扩容量能得到的最大权值可以同过二分枚举答案,用树链剖分判断。

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pb push_back
#define mp make_pair
#define Maxn 200010
#define Maxm 400010
#define LL __int64
#define Abs(x) ((x)>0?(x):(-x))
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define inf 100000
#define lowbit(x) (x&(-x))
#define clr(x,y) memset(x,y,sizeof(x))
#define Mod 1000000007
using namespace std;
int head[Maxn],vi[Maxn],dep[Maxn],w[Maxn],top[Maxn],son[Maxn],sz[Maxn],fa[Maxn],e,id;
int val[Maxn];
int num[Maxn],cnt;
struct Point{
    int val,i;
    int operator <(const Point &temp) const{
        return val<temp.val;
    }
}lis[Maxn];
struct Edge{
    int u,v,next;
    int val;
}edge[Maxn*3];
struct Tree{
    int l,r,c;
    int *p;
    int *q;
    int mid(){
        return (l+r)>>1;
    }
}tree[Maxn*3];
void init()
{
    clr(head,-1);clr(vi,0);
    e=0;id=0;
}
void add(int u,int v,int val)
{
    edge[e].u=u,edge[e].v=v,edge[e].val=val,edge[e].next=head[u],head[u]=e++;
}
void BuildTree(int l,int r,int po)
{
    tree[po].l=l,tree[po].r=r,tree[po].c=0;
    tree[po].p=new int[r-l+2];
    tree[po].q=new int[r-l+2];
    tree[po].q[0]=0;
    if(l==r)
        return ;
    int mid=tree[po].mid();
    BuildTree(l,mid,lson(po));
    BuildTree(mid+1,r,rson(po));
}
void update(int i,int c,int po)
{
    if(tree[po].l==tree[po].r){
        tree[po].p[++tree[po].c]=c;
        tree[po].q[tree[po].c]=tree[po].q[tree[po].c-1]+c;
        return ;
    }
    tree[po].p[++tree[po].c]=c;
    tree[po].q[tree[po].c]=tree[po].q[tree[po].c-1]+c;
    int mid=tree[po].mid();
    if(i<=mid)
        update(i,c,lson(po));
    else
        update(i,c,rson(po));
}
int getmin(int l,int r,int po)
{
    if(l<=tree[po].l&&tree[po].r<=r){
        return tree[po].p[1];
    }
    int mid=tree[po].mid();
    if(r<=mid)
        return getmin(l,r,lson(po));
    else if(l>=mid+1)
        return getmin(l,r,rson(po));
    else{
        return min(getmin(l,mid,lson(po)),getmin(mid+1,r,rson(po)));
    }
}
LL getans(int l,int r,LL val,int po)
{
    if(l<=tree[po].l&&tree[po].r<=r){
        int pos=lower_bound(tree[po].p+1,tree[po].p+tree[po].c+1,val)-tree[po].p;
        if(pos>tree[po].c){
            cnt+=tree[po].c;
            return tree[po].q[tree[po].c];
        }
        if(pos>1){
            cnt+=pos-1;
            return tree[po].q[pos-1];
        }
        return 0;
    }
    int mid=tree[po].mid();
    if(r<=mid)
        return getans(l,r,val,lson(po));
    else if(l>=mid+1)
        return getans(l,r,val,rson(po));
    else{
        return getans(l,mid,val,lson(po))+getans(mid+1,r,val,rson(po));
    }
}
void dfs(int u,int val)
{
    vi[u]=1;
    int i,v;
    son[u]=0,sz[u]=1;
    num[u]=val;
    for(i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(vi[v]) continue;
        dep[v]=dep[u]+1;
        fa[v]=u;
        dfs(v,edge[i].val);
        if(sz[v]>sz[son[u]])son[u]=v;
        sz[u]+=sz[v];
    }
}
void build(int u,int ti)
{
    int i,v;
    w[u]=++id;top[u]=ti;vi[u]=1;
    lis[id].i=id,lis[id].val=num[u];
    if(son[u]) build(son[u],ti);
    for(i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].v;
        if(vi[v]||v==son[u]) continue;
        build(v,v);
    }
}
LL need(LL val,int u,int v)
{
    int f1=top[u],f2=top[v];
    cnt=0;
    LL ans=0;
   // cout<<u<<" "<<f1<<" "<<v<<" "<<f2<<" "<<w[f1]<<" "<<w[u]<<endl;
    while(f1!=f2){
        if(dep[f1]<dep[f2]){
            swap(f1,f2),swap(u,v);
        }
        ans+=getans(w[f1],w[u],val,1);
        u=fa[f1];f1=top[u];
    }
    if(dep[u]>dep[v])
        swap(u,v);
    if(u!=v)
    ans+=getans(w[son[u]],w[v],val,1);
    //cout<<u<<" "<<v<<" "<<val<<" "<<cnt<<" "<<ans<<" "<<val*cnt-ans<<endl;
    ans=val*(LL)cnt-ans;
    return ans;
}
void calc(int u,int v,LL k,LL a,LL b)
{
    int x=u,y=v;
    int f1=top[u],f2=top[v];
    LL ans=10000000;
    while(f1!=f2){
        if(dep[f1]<dep[f2]){
            swap(f1,f2),swap(u,v);
        }
        ans=min(ans,(LL)getmin(w[f1],w[u],1));
        u=fa[f1];f1=top[u];
    }
    if(dep[u]>dep[v])
        swap(u,v);
    if(u!=v)
    ans=min(ans,(LL)getmin(w[son[u]],w[v],1));
    LL capacity=0;
    if(k<min(a,b)){
        printf("%I64d\n",ans);
        return ;
    }
    if(a<=b){
        printf("%I64d\n",k/a+ans);
        return ;
    }
    if(k>=a)
    capacity=(k-a)/b+1+ans;
    LL l,r,mid,temp;
    l=0,r=k/b+ans+1;
    while(l+1<r){
        mid=(l+r)>>1;
        temp=need(mid,x,y);
        temp*=b;
        if(temp<=k)
            l=mid;
        else
            r=mid-1;
    }
    if(r>=1)
    if(need(r,x,y)*b<=k&&r>l)
        l=r;
    printf("%I64d\n",max(capacity,l));
    return ;
}
int main()
{
    int t,n,m,i,j,u,v,val,q,a,b,k,Case=0;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d%d",&n,&m);
        for(i=1;i<n;i++){
            scanf("%d%d%d",&u,&v,&val);
            add(u,v,val);
            add(v,u,val);
        }
        dfs(1,1200000000);
        memset(vi,0,sizeof(vi));
        build(1,1);
        sort(lis+1,lis+id+1);
        BuildTree(1,n,1);
        for(i=1;i<=n;i++){
            update(lis[i].i,lis[i].val,1);
        }
        printf("Case #%d:\n",++Case);
        for(i=1;i<=m;i++){
            scanf("%d%d%d%d%d",&u,&v,&k,&a,&b);
            calc(u,v,k,a,b);
        }
    }
    return 0;
}

 

posted @ 2013-09-16 23:27 fangguo 阅读(...) 评论(...) 编辑 收藏