2017多校第9场 HDU 6162 Ch’s gift 树剖加主席树

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6162

题意：给出一棵树的链接方法，每个点都有一个数字，询问U-》V节点经过所有路径中l < = x < = r的数字和

解法：主席树维护区间和，树剖查询，复杂度nloglog。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
const int maxm = 40*maxn;
typedef long long LL;
int T[maxn];
int lson[maxm],rson[maxm];
LL sum[maxm];
int clk,M;
int newnode(){
    clk++;
    lson[clk]=rson[clk]=sum[clk]=0;
    return clk;
}
void update(int &now, int pre, int L, int R, int pos, int val)
{
    now = newnode();
    lson[now] = lson[pre];
    rson[now] = rson[pre];
    sum[now] = sum[pre] + val;
    if(L!=R){
        int mid=(L+R)>>1;
        if(pos<=mid) update(lson[now],lson[pre],L,mid,pos,val);
        else update(rson[now],rson[pre],mid+1,R,pos,val);
    }
}
LL query(int rt, int L, int R, int l, int r){
    if(l<=L&&R<=r){
        return sum[rt];
    }
    else{
        int mid = (L+R)>>1;
        LL ret = 0;
        if(l <= mid) ret += query(lson[rt], L, mid, l, r);
        if(mid < r) ret += query(rson[rt], mid+1, R, l, r);
        return ret;
    }
}
struct edge{
    int to,next;
}E[maxn*2];
int head[maxn],edgecnt,tim;
int siz[maxn],top[maxn],son[maxn],dep[maxn];
int fa[maxn],tid[maxn],Rank[maxn],val[maxn];
void init(){
    edgecnt=tim=0;
    memset(head,-1,sizeof(head));
    memset(son,-1,sizeof(son));
    clk=M=0;
    lson[clk]=rson[clk]=sum[clk]=0;
}
void add(int u, int v){
    E[edgecnt].to=v,E[edgecnt].next=head[u],head[u]=edgecnt++;
}
void dfs1(int u, int pre, int d){
    dep[u]=d;
    fa[u]=pre;
    siz[u]=1;
    for(int i=head[u];~i;i=E[i].next){
        int v=E[i].to;
        if(v!=pre){
            dfs1(v,u,d+1);
            siz[u]+=siz[v];
            if(son[u]==-1||siz[v]>siz[son[u]]) son[u]=v;
        }
    }
}
void dfs2(int u, int tp){
    top[u]=tp;
    tid[u]=++tim;
    Rank[tid[u]]=u;
    if(son[u]==-1) return;
    dfs2(son[u],tp);
    for(int i=head[u];~i;i=E[i].next){
        int v=E[i].to;
        if(v!=son[u]&&v!=fa[u]){
            dfs2(v,v);
        }
    }
}
int LCA(int u, int v)
{
    int ret;
    while(true)
    {
        if(top[u] == top[v])
        {
            ret = dep[u] < dep[v] ? u : v;
            break;
        }
        else if(dep[top[u]] > dep[top[v]])
            u = fa[top[u]];
        else v = fa[top[v]];
    }
    return ret;
}
int t[maxn];
LL query(int u, int v, int a, int b)
{
    a = lower_bound(t + 1, t + 1 + M, a) - t - 1;
    b = upper_bound(t + 1, t + 1 + M, b) - t - 1;
    if(b == 0)
        return 0;
    LL ret = 0;
    while(top[u] != top[v])
    {
        if(dep[top[u]] < dep[top[v]])
            swap(u, v);
        ret += query(T[tid[u]], 1, M, 1, b);
        ret -= query(T[tid[top[u]] - 1], 1, M, 1, b);
        if(a)
        {
            ret -= query(T[tid[u]], 1, M, 1, a);
            ret += query(T[tid[top[u]] - 1], 1, M, 1, a);
        }
        u = fa[top[u]];
    }
    if(dep[u] < dep[v])
        swap(u, v);
    ret += query(T[tid[u]], 1, M, 1, b);
    ret -= query(T[tid[v] - 1], 1, M, 1, b);
    if(a)
    {
        ret -= query(T[tid[u]], 1, M, 1, a);
        ret += query(T[tid[v] - 1], 1, M, 1, a);
    }
    return ret;
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        init();
        for(int i=1; i<=n; i++){
            scanf("%d", &val[i]);
            t[i]=val[i];
        }
        for(int i=1; i<n; i++){
            int u,v;
            scanf("%d %d", &u,&v);
            add(u, v);
            add(v, u);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        sort(t+1,t+n+1);
        M = unique(t+1, t+n+1)-t-1;
        for(int i=1; i<=n; i++) val[i] = lower_bound(t+1, t+1+M, val[i])-t;
        for(int i=1; i<=n; i++){
            T[i] = T[i-1];
            update(T[i],T[i],1,M,val[Rank[i]],t[val[Rank[i]]]);
        }
        for(int i=1; i<=m; i++){
            int u,v,x,y;
            scanf("%d %d %d %d", &u,&v,&x,&y);
            LL ret = query(u,v,x,y);
            printf("%lld", ret);
            if(i == m){
                printf("\n");
            }
            else{
                printf(" ");
            }
        }
    }
    return 0;
}