HDU 4366(Successor-块状链表)

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给一棵树(2<=n<=50000)个节点，每个节点有2个点权”忠诚”与”能力”
所有节点”忠诚”各不相同，现在有m个询问，每次对第i个节点，求出其子树中”能力”>它的节点中”忠诚”最高的那个

先求出dfs序，分n√段
每段按能力排序，预处理每段到i到末尾能取的最大忠诚
复杂度O(nn√logn)

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<map>#include<iomanip> #include<vector>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#pragma comment(linker, "/STACK:1024000000,1024000000") #define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>typedef long long ll;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (55000+10)int n,m;vi e[MAXN];int l[MAXN],r[MAXN],ti[MAXN];pi a[MAXN];void dfs(int x,int &T) {    l[x]=++T;    ti[T]=x;    for(auto v:e[x]) {        dfs(v,T);    }    r[x]=T;}map<int,int> h;int blo,le;int cmp(int x,int y) {    if (x==-1||y==-1) return x<y;    return a[x].fi<a[y].fi;}int sorted[MAXN],ma[MAXN];int work(int l,int r,int x) {    int ans=r+1;    while (l<=r) {        int m=(l+r)>>1;        if (a[sorted[m]].fi <=x) l=m+1;        else r=m-1,ans=m;    }    return ans;}int main(){//  freopen("hdu4366_data.out","r",stdin);//  freopen("hdu4366.out","w",stdout);    int T=read();    while(T--) {        h.clear();        int n=read(),m=read();        For(i,n-1) {            int u=read();            e[u].pb(i);            a[i].se=read(); a[i].fi=read();            h[a[i].se]=i;        }        int t=0;        dfs(0,t);        blo=250;        le=t/blo;        For(i,t) sorted[i]=ti[i];        For(i,le)        {             sort(sorted+(i-1)*blo+1,sorted+i*blo+1,cmp);        }        ForD(i,le*blo) {            ma[i]=a[sorted[i]].se;            if (i%blo)                 ma[i]=max(ma[i],ma[i+1]);        }        while(m--) {            int x;            scanf("%d",&x);            int L=l[x]+1,R=r[x];            if (L>R) {printf("-1\n");continue;}            int ans=-1;            for(int i=L;i<=R;)                if ( i % blo == 1 && i + blo - 1 <= R ) {                    if (a[sorted[i+blo-1]].fi<=a[x].fi ) {                        i+=blo;                        continue;                    }                    if (a[sorted[i]].fi>a[x].fi) {                        ans=max(ans, ma[i] );                        i+=blo;                         continue;                    } //这个优化必选加 //                  int p=upper_bound(sorted+i+1,sorted+i+blo,x,cmp)-(sorted); //这个函数不快但可用                     int p=work(i,i+blo-1,a[x].fi);                     if (p!=i+blo) {                        ans=max(ans,ma[p]);                    }                     i+=blo;                } else {                    if (a[x].fi<a[ti[i]].fi ) ans=max(ans,a[ti[i]].se);                    i++;                }            printf("%d\n",ans==-1 ? ans:h[ans]);        }        Rep(i,n) e[i].clear();    }       return 0;}

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