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_pairusing 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;}