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
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;
}