NEU 11月组队赛D题 QUERY ON THE TREE

题意

给定一个树，每个点有个权值v，vi=v(fa)*i%20161119,m个询问u,k，问以u为根的子树中abs(vj - k)最小值。

connection

思路

先dfs一遍求出dfs序和每个节点的权值，将问题转化为求区间中大于k的最小值和小于k的最大值。离线线段树即可。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define pii pair<int ,int>
#define mp make_pair
using namespace std;
typedef struct node node;
typedef struct que que;
const int maxnum = 101000;
const int mod = 20161119;
struct node
{
    int to,next;
    /* data */
}edge[2*maxnum];
struct que
{
    int k,num,id,u;
    /* data */
}q[maxnum];
que q0[maxnum];
int head[maxnum],minn[4*maxnum],maxn[4*maxnum],dfs_clock,l[maxnum],r[maxnum],big[maxnum],sml[maxnum];
long long num[maxnum];
int n,cnt,qq;
void add(int a,int b){
    edge[cnt].to = b;
    edge[cnt].next = head[a];
    head[a] = cnt++;
}
int update0(int a,int b,int &c){
    if(a == -1 || b == -1) {
        c = a==-1?b:a;
        return 1;
    }
    return 0;
}

void build(int l,int r,int o){
    if(l == r) maxn[o] = minn[o] = -1;
    else{
        int mid = (l + r)/2;
        build(l,mid,2*o);
        build(mid + 1,r,2*o + 1);
        maxn[o] = minn[o] = -1;
    }
}

void pushup(int l,int r,int o){
    if(!update0(maxn[2*o],maxn[2*o+1],maxn[o])) maxn[o] = max(maxn[2*o],maxn[2*o + 1]);
    if(!update0(minn[2*o],minn[2*o+1],minn[o])) minn[o] = min(maxn[2*o],maxn[2*o + 1]);
}

void update(int l,int r,int o,int k,int pos){
    if(l == r){
        minn[o] = maxn[o] = k;
    }
    else{
        int mid = (l + r)/2;
        if(pos <= mid) update(l,mid,2*o,k,pos);
        else update(mid + 1,r,2*o + 1,k,pos);
        pushup(l,r,o);
    }
}

pii query(int l,int r,int o,int ql,int qr){
    if(l > r) return mp(-1,-1);
    if(ql <= l && qr >= r){
        return mp(maxn[o],minn[o]);
    }
    int mid = (l + r)/2;
    pii ans;
    if(qr <= mid) ans = query(l,mid,2*o,ql,qr);
    else if(ql > mid) ans = query(mid + 1,r,2*o+1,ql,qr);
    else{
        pii ans1 = query(l,mid,2*o,ql,mid);
        pii ans2 = query(mid + 1,r,2*o+1,mid + 1,qr);
        if(!update0(ans1.first,ans2.first,ans.first)) ans.first = max(ans1.first,ans2.first);
        if(!update0(ans1.second,ans2.second,ans.second)) ans.second = min(ans1.second,ans2.second);
    }
    return ans;
}
void dfs(int u,int fa){
    num[u] = (num[fa] * (long long)u) %mod;
    l[u] = ++dfs_clock;
    for(int i = head[u];i != -1;i=edge[i].next){
        if(edge[i].to != fa)dfs(edge[i].to,u);
    }
    r[u] = dfs_clock;
}
int juedui(int a){
    return a<0?-a:a;
}
int cmp0(que a,que b){
    return num[a.u] < num[b.u];
}
int cmp1(que a,que b){
    return a.k < b.k;
}
int cmp2(que a,que b){
    return num[a.u] > num[b.u];
}
int cmp3(que a,que b){
    return a.k > b.k;
}
int cmp4(que a,que b){
    return a.id < b.id;
}
int main(){
    while(scanf("%d",&n)!= EOF){
        memset(head,-1,sizeof(head));
        cnt = dfs_clock = 0;
        int f,t;
        num[0] = 1;
        for(int i =2 ;i <= n;i ++) {
            scanf("%d%d",&f,&t);
            add(f,t);
            add(t,f);
        }
        build(1,n,1);
        dfs(1,0);
        scanf("%d",&qq);
        for(int i = 1;i <= qq;i ++){
            scanf("%d%d",&q[i].u,&q[i].k);
            q[i].id = i;
        }
        for(int i = 1;i <= n;i ++) q0[i].u = i;
        sort(q+1,q+1+qq,cmp1);
        sort(q0+1,q0+1+n,cmp0);
        int now = 1;
        for(int i = 1;i <= qq;i ++){
            while(now <=n && num[q0[now].u] <= q[i].k){
                update(1,n,1,num[q0[now].u],l[q0[now].u]);
                now ++;
            }
            sml[q[i].id] = query(1,n,1,l[q[i].u],r[q[i].u]).first; 
        }
        sort(q+1,q+1+qq,cmp3);
        sort(q0+1,q0+1+n,cmp2);
        build(1,n,1);
        now = 1;
        for(int i = 1;i <= qq;i ++){
            while(now <= n && num[q0[now].u] >= q[i].k){
                update(1,n,1,num[q0[now].u],l[q0[now].u]);
                now ++;
            }
            big[q[i].id] = query(1,n,1,l[q[i].u],r[q[i].u]).second;
        }
        sort(q+1,q+1+qq,cmp4);
        for(int i = 1;i <= qq;i ++){
            if(sml[i] == -1 || big[i] == -1) printf("%d\n",sml[i] == -1?juedui(q[i].k - big[i]):juedui(q[i].k - sml[i]) );
            else printf("%d\n",min(juedui(q[i].k - sml[i]),juedui(q[i].k - big[i])) );
        }   
    }   
    return 0;
}