hdu 6162 离线树链剖分

题意：一个树，点权，查询x到y路径上 价值在[a,b]之间的点的权值和

思路：剖分模板题，首先要想到离线处理，第一步将查询分为query(x,y,0,b)-query(x,y,0,a)；接下来我们就发现了可以按照点权的大小以及查询a的大小离线处理，这样就相当于只查询query(x,y)。。。故直接上个剖分点权的模板

代码：

#include<bits/stdc++.h>
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1)
#define mid (tr[rt].l+(tr[rt].r-tr[rt].l)/2)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define MEM(x,y) memset(x,y,sizeof(x));
#define bug(x) cout<<"bug"<<x<<endl;
typedef long long ll;
const int maxn =1e5+100;
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef pair<int,pair<int,int> > pii;

struct TREE{ll sum,l,r,tag;}tr[maxn*4];
int pos[maxn],sz[maxn],top[maxn],son[maxn],
fa[maxn],id[maxn],deep[maxn],cnt=0;
vector<int> g[maxn];
//void pushup(int rt){tr[rt].sum=tr[ls].sum+tr[rs].sum;}

void build(int l,int r,int rt){
    tr[rt].l=l,tr[rt].r=r;
    if(l==r){tr[rt].sum=0;return ;}
    int midd=l+(r-l)/2;
    build(l,midd,ls),build(midd+1,r,rs);
    tr[rt].sum=tr[ls].sum+tr[rs].sum;
}
void pushdown(int rt){
    if(tr[rt].tag){
        tr[ls].tag=(tr[ls].tag+tr[rt].tag);
        tr[rs].tag=(tr[rs].tag+tr[rt].tag);
        tr[ls].sum=(tr[ls].sum+(tr[ls].r-tr[ls].l+1)*tr[rt].tag);
        tr[rs].sum=(tr[rs].sum+(tr[rs].r-tr[rs].l+1)*tr[rt].tag);
        tr[rt].tag=0;
    }
}

void update(int l,int r,int c,int rt){
    if(l<=tr[rt].l&&tr[rt].r<=r){
        tr[rt].sum=(tr[rt].sum+(c*(tr[rt].r-tr[rt].l+1)));
        tr[rt].tag+=c;
        return ;
    }
    pushdown(rt);
    if(l<=mid) update(l,r,c,ls);
    if(r>mid) update(l,r,c,rs);
    tr[rt].sum=tr[ls].sum+tr[rs].sum;
}

ll query(int l,int r,int rt){
    if(l<=tr[rt].l&&tr[rt].r<=r) return tr[rt].sum;
    pushdown(rt);
    ll ans=0;
    if(l<=mid) ans+=query(l,r,ls);
    if(r>mid) ans+=query(l,r,rs);
    return ans;
}

void dfs1(int x,int fat,int dep){
    deep[x]=dep,fa[x]=fat,sz[x]=1;
    for(int i=0;i<g[x].size();i++){
        int v=g[x][i];
        if(v!=fat){
            dfs1(v,x,dep+1);
            sz[x]+=sz[v];
            if(son[x]==-1||sz[son[x]]<sz[v]) son[x]=v;
        }
    }
}

void dfs2(int x,int tp){
    top[x]=tp;pos[x]=++cnt;id[pos[x]]=x;
    if(son[x]==-1) return ;
    dfs2(son[x],tp);
    for(int i=0;i<g[x].size();i++){
        int v=g[x][i];
        if(v!=fa[x]&&v!=son[x])
            dfs2(v,v);
    }
}

ll add(int t1,int t2,int c,int ok){
    ll u=t1,v=t2,ans=0;
    //bug("F")
    while(top[u]!=top[v]){
        if(deep[top[u]]>deep[top[v]]) swap(u,v);
        if(!ok)update(pos[top[v]],pos[v],c,1);
        else ans+=query(pos[top[v]],pos[v],1);
        v=fa[top[v]];
    }
    if(deep[u]>deep[v]) swap(u,v);//!!
    if(!ok) update(pos[u],pos[v],c,1);
    else ans+=query(pos[u],pos[v],1);
    return ans;
}

int a[maxn];
pii Q[maxn*3];
int ql[maxn],qr[maxn],qll[maxn],qrr[maxn];
ll ans[maxn];
int main(){
    int n,P,v,u;
    while(scanf("%d%d",&n,&P)!=EOF){
        memset(son,-1,sizeof(son));
        MEM(ans,0);
        for(int i=0;i<=n+5;i++) g[i].clear();
        int x,y,num=0;
        cnt=0;
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
            Q[num++]=MP(a[i],MP(0,i));
        }
        for(int i=1;i<n;i++){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs1(1,-1,1);
        dfs2(1,1);
        build(1,n,1);
        for(int i = 1; i <=P; i++){
            scanf("%d%d%d%d",&ql[i],&qr[i],&qll[i],&qrr[i]);
            Q[num++]=MP(qll[i],MP(-1,i));
            Q[num++]=MP(qrr[i],MP(1,i));
        }
        sort(Q,Q+num);
        //bug(1)
        for(int i=0;i<num;i++){
            int ii=Q[i].Y.Y;
            if(Q[i].Y.X==0)add(ii,ii,Q[i].X,0);
            else if(Q[i].Y.X==-1)ans[ii]-=add(ql[ii],qr[ii],0,1);
            else ans[ii]+=add(ql[ii],qr[ii],0,1);
        }
        for(int i = 1; i <=P; i++)
            printf("%lld%c",ans[i]," 
"[i==P]);
    }
    return 0;
}