【HDU 6162】 Ch’s gift

【题目链接】

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

【算法】

          离线树剖

          我们知道，u到v路径上权值为[A,B]的数的和 = u到v路径上权值小于等于B的数的和 - u到v路径上权值小于等于(A-1)的数的和

          不妨将询问拆开，离线计算答案即可

【代码】

           

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;

int i,n,m,u,v,now,timer,tot,cnt;
long long l,r;
int head[MAXN],size[MAXN],fa[MAXN],top[MAXN],
        dep[MAXN],son[MAXN],dfn[MAXN];
long long ans[MAXN];

struct Edge
{
        int to,nxt;
} e[MAXN<<1];
struct info
{
        int val,pos;
} a[MAXN];
struct Query
{
        int u,v;
        long long m;
        int flag,id;    
} q[MAXN<<1];

struct SegmentTree
{
        struct Node
        {
                int l,r;
                long long sum;
        } Tree[MAXN<<2];
        inline void build(int index,int l,int r)
        {    
                int mid;
                Tree[index] = (Node){l,r,0};
                if (l == r) return;
                mid = (l + r) >> 1;
                build(index<<1,l,mid);
                build(index<<1|1,mid+1,r);
        }
        inline void update(int index)
        {
                Tree[index].sum = Tree[index<<1].sum + Tree[index<<1|1].sum;
        }
        inline void add(int index,int pos,long long val)
        {
                int mid;
                if (Tree[index].l == Tree[index].r) 
                {
                        Tree[index].sum += val;
                        return;
                }
                mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= pos) add(index<<1,pos,val);
                else add(index<<1|1,pos,val);
                update(index);
        }
        inline long long query(int index,int l,int r)
        {
                int mid;
                if (Tree[index].l == l && Tree[index].r == r) return Tree[index].sum;
                mid = (Tree[index].l + Tree[index].r) >> 1;
                if (mid >= r) return query(index<<1,l,r);
                else if (mid + 1 <= l) return query(index<<1|1,l,r);
                else return query(index<<1,l,mid) + query(index<<1|1,mid+1,r);
        }
} T;
inline bool cmp1(info a,info b)
{
        return a.val < b.val;        
}
inline bool cmp2(Query a,Query b)
{
        return a.m < b.m;
}
inline void add(int u,int v)
{
        tot++;
        e[tot] = (Edge){v,head[u]};
        head[u] = tot;
}
inline void dfs1(int u)
{
        int i,v;
        size[u] = 1;
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                if (fa[u] != v)
                {
                        fa[v] = u;
                        dep[v] = dep[u] + 1;
                        dfs1(v);
                        size[u] += size[v];
                        if (size[v] > size[son[u]]) son[u] = v;        
                }        
        }        
}
inline void dfs2(int u,int tp)
{
        int i,v;
        top[u] = tp;
        dfn[u] = ++timer;
        if (son[u]) dfs2(son[u],tp);
        for (i = head[u]; i; i = e[i].nxt)
        {
                v = e[i].to;
                if (fa[u] != v && son[u] != v) dfs2(v,v);
        }
}
inline long long query(int u,int v)
{
        int tu = top[u],
                tv = top[v];
        long long ret = 0;
        while (tu != tv)
        {
                if (dep[tu] > dep[tv])
                {
                        swap(u,v);
                        swap(tu,tv);
                }
                ret += T.query(1,dfn[tv],dfn[v]);
                v = fa[tv]; tv = top[v];
        }
        if (dfn[u] > dfn[v]) swap(u,v);
        ret += T.query(1,dfn[u],dfn[v]);
        return ret;
}

int main() 
{
        
        while (scanf("%d%d",&n,&m) != EOF)
        {
                timer = 0;
                tot = 0;
                for (i = 1; i <= n; i++) 
                {
                        head[i] = 0;
                        son[i] = 0;
                }
                cnt = 0;
                memset(ans,0,sizeof(ans));
                for (i = 1; i <= n; i++) 
                {
                        scanf("%lld",&a[i].val);
                        a[i].pos = i;
                }
                for (i = 1; i < n; i++)
                {
                        scanf("%d%d",&u,&v);
                        add(u,v);
                        add(v,u);
                }
                dfs1(1);
                dfs2(1,1);
                T.build(1,1,timer);
                for (i = 1; i <= m; i++)
                {
                        scanf("%d%d%lld%lld",&u,&v,&l,&r);
                        if (l > 1) q[++cnt] = (Query){u,v,l-1,1,i};
                        q[++cnt] = (Query){u,v,r,2,i};
                }
                sort(a+1,a+n+1,cmp1);
                sort(q+1,q+cnt+1,cmp2);
                now = 1;
                for (i = 1; i <= cnt; i++)
                {
                        while (now <= n && a[now].val <= q[i].m)
                        {
                                T.add(1,dfn[a[now].pos],a[now].val);
                                now++;
                        }    
                        if (q[i].flag == 1) ans[q[i].id] -= query(q[i].u,q[i].v);
                        else ans[q[i].id] += query(q[i].u,q[i].v);
                }
                for (i = 1; i < m; i++) printf("%lld ",ans[i]);
                printf("%lld\n",ans[m]);
        }
        
        return 0;
    
}