BZOJ-2588: Spoj 10628. Count on a tree（树上路径第K最值=LCA+可持久化线段树）

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2588

思路：

每个节点上建立一棵维护权值的可持久化线段树（维护从根到这个节点的权值），以他的父节点为历史版本建立，每次查询时直接在线段树上二分即可，所以只需要联立三棵可持久化线段树T[u],T[v],T[lca(u,v)]即可快捷查询。复杂度O(n log n)****

****代码：****

#include 

#include 

#include 

#include 

  

using namespace std;

  

#define MAXN 100001

  

int a[MAXN];

int n,m;

  

struct edge {

    int t;

    bool flag;

    edge *next;

    edge (){

        next=NULL;

        flag=false;

    }

} *head[MAXN];

  

void INSERT(int s,int t){

    edge *p=new(edge);

    p->t=t;

    p->next=head[s];

    head[s]=p;

}

  

struct Node {

    int l,r,mint;

} RMQ[MAXN*6];

  

int DFS[MAXN*2],V=0;

int h[MAXN],first[MAXN];

bool f[MAXN];

  

void dfs(int v){

    f[v]=false;

    DFS[++V]=v;

    first[v]=V;

    for (edge *p=head[v];p;p=p->next){

        if (f[p->t]){

            p->flag=true;

            h[p->t]=h[v]+1;

            dfs(p->t);

            DFS[++V]=v;

        }

    }

}

  

void Build_RMQ(int l,int r,int t){

    RMQ[t].l=l;

    RMQ[t].r=r;

    if (l==r){

        RMQ[t].mint=DFS[l];

        return ;

    }

    int mid=(l+r)>>1;

    Build_RMQ(l,mid,t<<1);

    Build_RMQ(mid+1,r,(t<<1)^1);

    if (h[RMQ[t<<1].mint]>1;

    if (r<=mid) return query(l,r,t<<1);

    if (l>mid) return query(l,r,(t<<1)^1);

    int L=query(l,mid,t<<1),R=query(mid+1,r,(t<<1)^1);

    if (h[L]l=l;

    t->r=r;

    if (l==r) return ;

    Build0(l,(l+r)>>1,t->left=new(node));

    Build0(((l+r)>>1)+1,r,t->right=new(node));

}

  

  

void Add(int x,node *u,node *t){

    t->l=u->l;

    t->r=u->r;

    t->s=u->s+1;

    if (t->l==t->r) return ;

    int mid=(t->l+t->r)>>1;

    if (x<=mid) {

        t->right=u->right;

        t->left=new(node);

        Add(x,u->left,t->left);

    } else {

        t->left=u->left;

        t->right=new(node);

        Add(x,u->right,t->right);

    }

}

  

queueQ;

  

void Build(){

    Build0(1,N,T[0]=new(node));

    INSERT(0,1);

    head[0]->flag=true;

    while (!Q.empty()) Q.pop();

    Q.push(0);

    while (!Q.empty()) {

        int v=Q.front();

        Q.pop();

        for (edge *p=head[v];p;p=p->next){

            if (p->flag){

                Add(r[p->t],T[v],T[p->t]=new(node));

                Q.push(p->t);

            }

        }

    }

}

  

int main(){

    scanf("%d%d",&n,&m);

    for (int i=0;i++l!=s->r) {

            int sum=s->left->s+t->left->s-2*l->left->s;

            if (a[L]<=c[(s->l+s->r)>>1]&&a[L]>=c[s->l]) sum++;

            if (k<=sum) {

                s=s->left;

                t=t->left;

                l=l->left;

            } else {

                k-=sum;

                s=s->right;

                t=t->right;

                l=l->right;

            }

        }

        lastans=c[s->l];

        printf("%d",lastans);

        if (m) printf("\n");

    }

    return 0;

}