acm数据结构整理

 一、区间划分
 //区间划分+持久化并查集：区间连通情况统计。
 inline bool comp(Ask x, Ask y){return x.km == y.km ? x.l > y.l : x.km > y.km ; }
 inline void init()//把编号相同的放一起，l从大到小；{
     int i ;
     ms = sqrt(1.0*n) ; //大小
     mg = n/ms ; //块数
     mg = mg*ms == n ? mg : mg +  ;
     for(i =  ; i <= n ; i ++) e[i].clear() ;
 }
 /*********************持久化并查集**********************************/
 void solve()//区间划分{
     int i = , j, k, l, r ;
     sort(ask+, ask++q, comp) ;
     while(i <= q) //询问
     {
         k = ask[i].km ;
         for(; i <= q && ask[i].km == k && ask[i].l > (k-)*ms ; i ++) //处理区间小于ms的询问
             ans[ask[i].no] = bfAnswer(i) ;
         if(i > q || ask[i].km != k ) //全部为小区间
             continue ;
         r = (k-)*ms ; //最右端
         initSet(lp, , r) ; //初始化并查集 ；
         for(j =  ; j <= r ; j ++) vis[j] =  ;
         while(i <= q && ask[i].km == k ) //求解第k块的所有大询问；
         {
             l = ask[i].l ;
             for(j = r ; j >= l ; j -- ) //加入顶点j
                 addVex(lp, j, (k-)*ms) ;
             r = j ;
             for(; i <= q && ask[i].km == k && ask[i].l == l ; i ++)
 //计算所有左端等于l的区间；
             {
                 ans[ask[i].no] = unionRSet(ask[i].no, l, (k-)*ms+, ask[i].r ) ; //右端暴力合并 ；
             }
         }
     }
 }
 二、2B树：动态区间比x小的数的个数。
 void DT_insert(int rt, int x, int s ){
     int i = DN, j, p ;
     getBinary(x) ;
     j = dig[i--] ;
     if(root[rt].next[j] == -)
     {
         root[rt].next[j] = ++ tot ;
         pool[tot].set() ;
     }
     if(!j) root[rt].sum += s ;
     p = root[rt].next[j] ;
     while(i)
     {
         j = dig[i--] ;
         if(pool[p].next[j] == -)
         {
             pool[p].next[j] = ++ tot ;
             pool[tot].set() ;
         }
         if(!j)pool[p].sum += s ;
         p = pool[p].next[j] ;
     }
 }
 int DT_count(int rt, int x){
     int i = DN, j, p, ans =  ;
     getBinary(x) ;
     j = dig[i--] ;
     if(j) ans += root[rt].sum ;
     if(root[rt].next[j] == -) return ans ;
     p = root[rt].next[j] ;
     while(i)
     {
         j = dig[i--] ;
         if(j) ans += pool[p].sum ;
         if(pool[p].next[j] == -) return ans ;
         p = pool[p].next[j] ;
     }
     return ans ;
 }
 void DT_build(){
     int i, k ;
     size = sqrt(double(n)) ;
     mg = n/size ;
     mg = mg*size == n ? mg : mg+ ;
     for(i =  ; i <= mg ; i ++) root[i].set() ;
     for(i =  ; i <= n ; i ++)
     {
         k = (i-)/size +  ;
         DT_insert(k, ar[i], ) ;
     }
     bs = *size ;
     bmg = n/bs ;
     bmg = bmg*bs == n ? bmg : bmg +  ;
     for(i =  ; i <= bmg+ ; i ++) root[MAXS+i].set() ;
     for(i =  ; i <= n ; i ++)
     {
         k = (i-)/bs +  ;
         DT_insert(MAXS+k, ar[i], ) ;
     }
 }
 int DT_query(int l, int r, int x){
     int i, k, ans = , bk ;
     k = (l-)/size +  ;
     k = size*(k-)+ == l ? k : k +  ;  //前
 for(i = l ; i <= r && i < (k-)*size+ ; i ++) ans = ar[i] < x ? ans+ : ans ;
 for(; i <= r && r-i+ >= size && (k% != ) ; i += size, k ++) //小块
         ans += DT_count(k, x) ;
 for(bk = (k-)/+ ; i <= r && r-i+ >= bs ; i += bs, bk ++, k += )
      ans += DT_count(MAXS+bk, x) ;//大块
     for(; i <= r && r-i+ >= size ; i += size, k ++) //小块
         ans += DT_count(k, x) ;
     for(; i <= r ; i ++) ans = ar[i] < x ? ans+ : ans ;
     return ans ;
 }
 三、树链剖分
 . 树上区间[u, v] 异或 x 的最大值
 // 持久化2B树 + 树链剖分（区间可直接合并）
 inline void insert(int rt, int x) ； //建一颗空树
 inline void addNumber(int pre, int &rt, int id, int x){//持久化
     int i = SD, p ;
     bool j ;
     rt = ++dcnt ; p = rt ;
     while(i >=  )
     {
         pool[p] = pool[pre] ;
         j = x&(<<i) ;
         pool[p].mx[j] = id ;
         pool[p].next[j] = ++dcnt ;
         p = dcnt ;
         pre = pool[pre].next[j] ;
         i -- ;
     }
     pool[p].x = x ;
 }
 inline int DTcount(int rt,int l,int x)//计算前缀树[1,r]中下标>=l的与x的最大异或值
 /*******c******************树链剖分*************************/
 void DFS(int u){
     int i, to ;
     siz[u] =  ; son[u] =  ;
     for(i = head[u] ; i != - ; i = e[i].next )
     {
         to = e[i].to ;
         if(to != fa[u])
         {
             fa[to] = u ;
             dep[to] = dep[u] +  ;
             DFS(to) ;
             if(siz[son[u]] < siz[to]) son[u] = to ;
             siz[u] += siz[to] ;
         }
     }
 }
 void cutLink(int u, int tp){
     w[u] = ++cnt ;
     top[u] = tp ;
     if(son[u]) cutLink(son[u], tp) ;
     for(int i = head[u] ; i != - ; i = e[i].next)
     {
         int to = e[i].to ;
         if(to != fa[u] && to != son[u]) cutLink(to, to) ;
     }
 }
 void buildTree(){
     dep[] =  ;
     cnt = fa[] = siz[] =  ;
     DFS() ;
     cutLink(, ) ;
     int i ;
     for(i =  ; i <= n ; i ++) val[w[i]] = ar[i] ;
     pool[].set() ;
     dcnt = T[] =  ;
     for(i =  ; i <= cnt ; i ++) insert(T[], val[i]) ;
     for(i =  ; i <= cnt ; i ++) addNumber(T[i-], T[i], i, val[i]) ;//持久化思
 }
 int getAns(int u, int v, int x){
     int ans = , tu = top[u], tv = top[v], tmp ;
     if(u == v) return DTcount(T[w[u]], w[v], x) ; ;
     while(tu != tv) //如果不在同一条链上
     {
         if(dep[tu] < dep[tv]) {swap(tu, tv) ; swap(u, v) ; }
         tmp = DTcount(T[w[u]], w[tu], x ) ; //查询从u到tu的父亲的最大值
         ans = max(ans, tmp) ;
         u = fa[tu] ;
         tu = top[u] ;
     }
     if(dep[u] < dep[v]) swap(u, v) ;
     tmp = DTcount(T[w[u]], w[v], x) ;
     return max(ans, tmp) ;
 }
 . BFS实现建树
 int leaf[MAXN], ns[MAXN], ct[MAXN];
 void BFS(int rt){
     int i, h = , t = , l = , r =  ;
     int u, to ;
     que[t++] = rt ;
     for(i =  ; i <= n ; i ++)
     {
         siz[i] =  ;
         ct[i] = ns[i] = son[i] =  ;
     }
     while (h < t) // calculate the dep.
     {
         u = que[h++] ;
         for(i = head[u] ; i != - ; i = e[i].next)
         {
             to = e[i].to ;
             if(to != fa[u])
             {
                 ns[u] ++ ;
                 fa[to] = u ;
                 dep[to] = dep[u] +  ;
                 que[t++] = to ;
             }
         }
     }
     for(i =  ; i <= n ; i ++) if(ns[i] == ) leaf[r++] = i ;
     while(l < r)
     {
         to = leaf[l++] ;
         u = fa[to] ;
         siz[u] += siz[to] ; ct[u] ++ ;
         if(ct[u] == ns[u] && u ) leaf[r++] = u ;
     }
     h = t =  ;
     que[t++] = rt ;
     siz[] =  ;
     while (h < t)
     {
         u = que[h++] ;
         for(i = head[u] ; i != - ; i = e[i].next)
         {
             to = e[i].to ;
             if(to != fa[u])
             {
                 if(siz[son[u]] < siz[to]) son[u] = to ;
                 que[t++] = to ;
             }
         }
     }
 }
 void BFS_cutLink(int rt, int tp){
     int i, u, to, h = , t =  ;
     que[t++] = rt ;
     while (h < t)
     {
         u = que[h++] ;
         w[u] = ++cnt ; top[u] = u ;
         to = son[u] ;
         while (to)
         {
             w[to] = ++cnt ; top[to] = u ;
             for(i = head[to] ; i != - ; i = e[i].next)
                 if(e[i].to != fa[to] && e[i].to != son[to]) que[t ++] = e[i].to ;
             to = son[to] ;
         }
         for(i = head[u] ; i != - ; i = e[i].next)
         {
             to = e[i].to ;
             if(to != fa[u] && to != son[u]) que[t++] = to ;
         }
     }
 }
 . 非直接合并查询：树上最长上升子序列
 /*********************线段树成段更新最长上升子序列*************************/
 void Up(int step){
 L[step].lx = L[lson].lx + ( (L[lson].lx==L[lson].len && val[L[rson].left] > val[L[lson].right])?L[rson].lx:) ;
     L[step].rx = L[rson].rx + ((L[rson].rx==L[rson].len &&val[L[rson].left] > val[L[lson].right])?L[lson].rx:) ;
 L[step].mx = max(max(L[lson].mx,L[rson].mx),val[L[rson].left]
  > val[L[lson].right]?(L[lson].rx+L[rson].lx):);
 
    L[step].dlx = L[lson].dlx + ( (L[lson].dlx == L[lson].len && val[L[rson].left] < val[L[lson].right]) ? L[rson].dlx :  );
     L[step].drx = L[rson].drx + ((L[rson].drx == L[rson].len && val[L[rson].left] < val[L[lson].right]) ? L[lson].drx :  ) ;
 L[step].dmx = max(max(L[lson].dmx,L[rson].dmx),val[L[rson].left]
  < val[L[lson].right] ? (L[lson].drx+L[rson].dlx) :  ) ;
 }
 void build(int step,int l,int r){
     L[step].left=l;
     L[step].right=r;
     L[step].len = r-l+ ;
     L[step].mid=(l+r)/;
     if(l==r)
     {
         L[step].lx = L[step].rx = L[step].mx =  ;
         L[step].dlx =L[step].drx = L[step].dmx =  ;
         return ;
 }
 …… ；
 }
 inline void unionNode(int l, int r, int ar ){
 L[ar].lx = L[l].lx + ((L[l].lx==L[l].len && val[L[r].left]
 > val[L[l].right]) ? L[r].lx : ) ;
 L[ar].rx = L[r].rx + ((L[r].rx==L[r].len && val[L[r].left]
 > val[L[l].right]) ? L[l].rx : ) ;
 L[ar].mx = max(max(L[l].mx,L[r].mx),val[L[r].left]
 > val[L[l].right]?(L[l].rx+L[r].lx):);
 
 L[ar].dlx = L[l].dlx + ((L[l].dlx == L[l].len && val[L[r].left]
 < val[L[l].right]) ? L[r].dlx :  );
 L[ar].drx = L[r].drx + ((L[r].drx == L[r].len && val[L[r].left]
 < val[L[l].right]) ? L[l].drx :  ) ;
 L[ar].dmx = max(max(L[l].dmx,L[r].dmx),val[L[r].left] < val[L[l].right]
 ? (L[l].drx+L[r].dlx) :  ) ;
     L[ar].left = L[l].left ;
     L[ar].right = L[r].right ;
     L[ar].len = L[l].len + L[r].len ;
 }
 int query(int step,int l,int r){
     if(l==L[step].left&&r==L[step].right) return step ;
     …… ；
     else    {
         int lt = query(lson,l,L[step].mid);
         int rt = query(rson,L[step].mid+,r);
         ++scnt ;
         unionNode(lt, rt, scnt) ;
         return scnt ;
     }
 }
 /*************************树链剖分*************************/
 inline int unionAns(int l, int r){
     int tmp = max(L[l].dmx, L[r].mx), ans ;
     if(val[L[l].left] < val[L[r].left]) ans = L[r].lx + L[l].dlx ;
     else ans =  ;
     ans = max(tmp, ans) ;
     return ans ;
 }
 int getAns(int u, int v){
     if(u == v) return L[query(, w[u], w[u])].mx ;
     int tu = top[u], tv = top[v] ;
     int ar, tr, ur, vr ;
     scnt = *cnt +;
     ar = ++scnt ; ur = - ; vr = - ;
     while (tu != tv)
     {
         if(dep[tu] > dep[tv])//calculate the link tu -> u.
         {
             tr = query(, w[tu], w[u]) ;
             if(ur == -) ur = tr ;
             else            {
                 ++scnt ;
                 unionNode(tr, ur, scnt); //tr is left node.
                 ur = scnt ;
             }
             u = fa[tu] ; tu = top[u] ;
         }
         else        {
             tr = query(, w[tv], w[v]) ;
             if(vr == -) vr = tr ;
             else            {
                 ++scnt ;
                 unionNode(tr, vr, scnt) ; // tr is left node.
                 vr = scnt ;
             }
             v = fa[tv] ; tv = top[v] ;
         }
     }
     if(dep[u] >= dep[v])// v is root.
     {
         tr = query(, w[v], w[u]) ;
         if(ur == -) ur = tr ;
         else        {
             scnt ++ ;
             unionNode(tr, ur, scnt) ;
             ur = scnt ;
         }
         u = v ;
     }
     else    {
         tr = query(, w[u], w[v]) ;
         if(vr == -) vr = tr ;
         else        {
             ++scnt ;
             unionNode(tr, vr, scnt) ;
             vr = scnt ;
         }
         v = u ;
     }
     if(ur == -) return L[vr].mx ;
     else if(vr == -) return L[ur].dmx ;
     return unionAns(ur, vr) ; // calculate the answer of u->v
 }
 . 树上最大字段和
 /*********************线段树最大子段和*************************/
 inline void Up(int rt){
     sum[rt] = sum[rt<<] + sum[rt<<|] ;
     lx[rt] = max(lx[rt<<|], lx[rt<<] + sum[rt<<|]) ;
     rx[rt] = max(rx[rt<<], rx[rt<<|] + sum[rt<<]) ;
     mx[rt] = max(mx[rt<<], mx[rt<<|]) ;//不合并
     mx[rt] = max(mx[rt], lx[rt<<] + rx[rt<<|]) ; //合并
 }
 inline void Down(int rt, int L, int R){
     if(d[rt] != INF)
     {
         int x = d[rt] ;
         int mid = (L+R) >>  ;
         d[rt] = INF ;
         d[rt<<] = d[rt<<|] = x ;
         sum[rt<<] = (LL)x*(mid-L+) ;
         sum[rt<<|] = (LL)x*(R-mid) ;
         lx[rt<<] = rx[rt<<] = mx[rt<<] = x >  ? sum[rt<<] :  ;
         lx[rt<<|] = rx[rt<<|] = mx[rt<<|] = x >  ? sum[rt<<|] :  ;
     }
 }
 inline void unionNode(int l, int r, int ar ){
     sum[ar] = sum[l] + sum[r] ;
     lx[ar] = max(lx[r], lx[l]+sum[r]) ;
     rx[ar] = max(rx[l], rx[r]+sum[l]) ;
     mx[ar] = max(mx[l], mx[r]) ;
     mx[ar] = max(mx[ar], lx[l]+rx[r]) ;
 }
 void build(int rt, int l, int r){
     d[rt] = INF ;
     if(l == r)
     {
         sum[rt] = lx[rt] = rx[rt] = mx[rt] = val[l] ;
         return ;
 }
 …… ；
 }
 void updata(int rt, int L, int R, int l, int r, int x){
     if(L == l && r == R)
     {
         sum[rt] = (R-L+)*x ;
         lx[rt] = rx[rt] = mx[rt] = x >  ? sum[rt] :  ;
         d[rt] = x ;
         return ;
     }
 Down(rt, L, R) ;
 …… ；
 }
 int query(int rt, int L, int R, int l, int r){
     if(L == l && R == r) return rt ;
     Down(rt, L, R) ;
     …… ；
     else    {
         int x = query(rt<<, L, mid, l, mid) ;
         int y = query(rt<<|, mid+, R, mid+, r) ;
         ++scnt ;
         unionNode(x, y, scnt) ;
         return scnt ;
     }
 }
 /*************************树链剖分*************************/
 inline void unionAns(int x, int y, int ar){
     mx[ar] = max(mx[x], mx[y]) ;
     mx[ar] = max(mx[ar], rx[x]+rx[y]) ;
 }
 int getAns(int u, int v){
     if(u == v) return query(, , cnt, w[u], w[u]) ;
     int tu = top[u], tv = top[v] ;
     int ar, tr, ur, vr ;
     scnt = *cnt +;
     ar = ++scnt ; ur = - ; vr = - ;
     while (tu != tv)
     {
         if(dep[tu] > dep[tv])//calculate the link tu -> u.
         {
             tr = query(, , cnt, w[tu], w[u]) ;
             if(ur == -) ur = tr ;
             else            {
                 ++scnt ;
                 unionNode(tr, ur, scnt); //tr is left node.
                 ur = scnt ;
             }
             u = fa[tu] ; tu = top[u] ;
         }
         else        {
             tr = query(, , cnt, w[tv], w[v]) ;
             if(vr == -) vr = tr ;
             else            {
                 ++scnt ;
                 unionNode(tr, vr, scnt) ; // tr is left node.
                 vr = scnt ;
             }
             v = fa[tv] ; tv = top[v] ;
         }
     }
     if(dep[u] >= dep[v])
     {
         tr = query(, , cnt, w[v], w[u]) ;
         if(ur == -) ur = tr ;
         else        {
             scnt ++ ;
             unionNode(tr, ur, scnt) ;
             ur = scnt ;
         }
         u = v ;
     }
     else    {
         tr = query(, , cnt, w[u], w[v]) ;
         if(vr == -) vr = tr ;
         else        {
             ++scnt ;
             unionNode(tr, vr, scnt) ;
             vr = scnt ;
         }
         v = u ;
     }
     if(ur == -) return vr ;
     else if(vr == -) return ur ;
     unionAns(ur, vr, ar) ; // calculate the answer of u->v
     return ar ;
 }
 四、持久化线段树
 /*******************静态 Kth number ***************************************/
 void build(int &rt, int l, int r ){
     rt = ++tot ;
     sum[rt] =  ;
 if(l >= r) return ;
 …… ；
 }
 void updata(int &rt, int pre, int l, int r, int i, int v){
     rt = ++tot ;
     ls[rt] = ls[pre] ; rs[rt] = rs[pre] ; sum[rt] = sum[pre] + v ;
     if(l >= r ) return ;
     int mid = l+r >>  ;
     if(mid >= i ) updata(ls[rt], ls[pre], l, mid, i, v) ;
     else updata(rs[rt], rs[pre], mid+, r, i, v) ;
 }
 int query(int R, int L, int l, int r, int k){
     if(l >= r) return l ;
     int t = sum[ls[R]] - sum[ls[L]] ;
     int mid = l+r >>  ;
     if(t >= k) return query(ls[R], ls[L], l, mid, k) ;
     else return query(rs[R], rs[L], mid+, r, k-t) ;
 }
 /***************************************************************/
 void solve(){
      sort(ord+, ord++n, comp) ;
      cnt = rk[ord[]] =  ;
      nar[cnt] = ar[ord[]] ;
      for(i =  ; i <= n ; i ++)
      {
           if(ar[ord[i]] != ar[ord[i-]])
           {
               rk[ord[i]] = ++cnt ;
               nar[cnt] = ar[ord[i]] ;
           }
           else rk[ord[i]] = cnt ;
       }
       tot =  ;
       build(T[], , cnt) ;
       for(i =  ; i <= n ; i ++) updata(T[i], T[i-], , cnt, rk[i], ) ;
       for(i =  ; i <= m ; i ++)
       {
           scanf("%d %d %d", &l, &r, &k) ;
           ans = query(T[r], T[l-], , cnt, k) ;
           printf("%d\n", nar[ans]) ;//离散后的数组
       }
     }
   }
 五、KD树
 #define N 50005
 #define MID(x,y) ( (x + y)>>1 )
 using namespace std ;
 typedef long long LL ;
 struct Point{
     int x[];
     LL dis;
     Point(){
         for(int i =  ; i <  ; i++) x[i] =  ;
         dis = 9223372036854775807LL ;
     }
 friend bool operator < (const Point &a, const Point &b)
 { return a.dis < b.dis ; }
 };
 priority_queue <Point, vector<Point> > res ;
 LL dist2(const Point &a, const Point &b, int k) {
 //距离的平方,开根号很耗时，能不开就不开
     LL ans = ;
 for (int i = ; i < k; i++)
 //一开始这儿写的i < 5，WA了N次。。。原本以为只要初始值设0就无所谓,
         ans += (a.x[i] - b.x[i]) * (a.x[i] - b.x[i]);
    //但发现Point有个全局变量，在多case下会出错。。。
     return ans;
 }
 int ddiv;
 bool cmpX(const Point &a, const Point &b){return a.x[ddiv] < b.x[ddiv];}
 struct KDTree   {
     Point p[N];        //空间内的点
 int Div[N];
 //记录区间是按什么方式划分（分割线平行于x轴还是y轴, ==1平行y轴切;==0平行x轴切）
     int k;          //维数
     int m;          //近邻
 int getDiv(int l, int r)   {     //寻找区间跨度最大的划分方式
         map <int, int> ms;
         int minx[],maxx[];
         for (int i = ; i < k; i++)
         {
             ddiv = i;
             minx[i] = min_element(p + l, p + r + , cmpX)->x[i];
             maxx[i] = max_element(p + l, p + r + , cmpX)->x[i];
             ms[maxx[i] - minx[i]] = i;
         }
         map <int ,int>::iterator pm = ms.end();
         pm--;
         return pm->second;
     }
     void build(int l, int r)  {      //记得先把p备份一下。
         if (l > r)    return;
         int mid = MID(l,r);
         Div[mid] = getDiv(l,r);
         ddiv = Div[mid];
         nth_element(p + l, p + mid, p + r + , cmpX);
         build(l, mid - );
         build(mid + , r);
     }
     void findk(int l, int r, Point a) {//k(m)近邻，查找k近点的平方距离
         if (l > r)    return;
         int mid = MID(l,r);
         LL dist = dist2(a, p[mid], k);
         if (dist >= )
         {
             p[mid].dis = dist;
             res.push(p[mid]);
             while ((int)res.size() > m)
                 res.pop();
         }
         LL d = a.x[Div[mid]] - p[mid].x[Div[mid]];
         int l1, l2, r1, r2;
         l1 = l , l2 = mid + ;
         r1 = mid - , r2 = r;
         if (d > )
             swap(l1, l2), swap(r1, r2);
         findk(l1, r1, a);
         if ((int)res.size() < m || d*d < res.top().dis )
             findk(l2, r2, a);
     }
 };
 Point pp[N];
 KDTree kd;
 int solve(){
        for (int i = ; i < n; i++)
             for (int j = ; j < kd.k; j++)
             {
                 scanf("%d", &pp[i].x[j]);
                 kd.p[i] = pp[i];
             }
         kd.build(, n - );
         int t =  ;
         while(t--)
         {
             Point a;
             for (int i = ; i < kd.k; i++)scanf("%d", &a.x[i]);
             scanf("%d", &kd.m);
             kd.findk(, n - , a);
             printf("the closest %d points are:\n", kd.m);
             Point ans[];
             for (int i = ; !res.empty(); i++)
             {
                 ans[i] = res.top();
                 res.pop();
             }
             for (int i = kd.m - ; i >= ; i--)
             {
                 for (int j = ; j < kd.k - ; j++)
                     printf("%d ", ans[i].x[j]);
                 printf("%d\n", ans[i].x[kd.k - ]);
             }
         }
 六、最远manhattan距离
 int query(int rt, int p, int s) //维护每个状态的最大值、最小值。
 void getTable(int pos){//每个点有sta个状态
     int i, j, t ;
     for(i =  ; i < sta ; i ++)
     {
         t = i ;
         tmp[i] =  ;
         for(j =  ; j <= k ; j ++)
         {
             if(t&) tmp[i] += ar[pos].x[j] ;
             else tmp[i] -= ar[pos].x[j] ;
             t >>=  ;
         }
     }
 }
 int solve(){
         sta = <<k ; //K维平面
         build(, , q) ;//将每个点的sta个状态插入树中
         cnt =  ;
         for(i =  ; i <= q ; i ++)
         {
             if(cnt < ){puts("") ;continue ; }
             int ans =  ;
             for(j =  ; j < sta ; j ++)
             {
                 mxs = query(, j, ) ; //查询最大值；
                 mns = query(, j, ) ; //查询最小值；
                 ans = ans > mxs - mns ? ans : mxs-mns ;
             }
             printf("%d\n", ans) ;
         }
 七、LCA & RMQ & 树状数组
 /*****************************LCA*******************************/
 struct QNode{
     int from, to, next, no ;
 }qe[MAXN];
 int cnt ;
 int qhead[MAXN], pre[MAXN], LCA[MAXN] ;
 inline void addQEdge(int a,int b,int c){
     qe[++cnt].from = a ;
     qe[cnt].to = b ;
     qe[cnt].next = qhead[a] ;
     qe[cnt].no = c ;
     qhead[a] = cnt;
 }
 int find(int x){
     if(x != pre[x]) pre[x]=find(pre[x]);
     return pre[x];
 }
 void tarjan(int u){
     int j, v ;
     pre[u] = u ;
     for(j = qhead[u] ; j != - ; j = qe[j].next)
     {
         v = qe[j].to;
         LCA[qe[j].no] = find(v) ;
     }
     for(j = head[u] ; j != - ;j = e[j].next)
     {
         v = e[j].to ;
         tarjan(v) ;
         pre[v] = u ;
     }
 }
 /*****************************RMQ*******************************/
 int mn[MAXN][] ;
 void rmq_init(int num){
     int i , j ;
     for(j =  ; j <= num ; j ++ ) mn[j][] = height[j] ;
     int m = floor(log((double)num)/log(2.0)) ;
     for(i =  ; i <= m ; i ++ )
     {
         for(j = num ; j >  ; j -- )
         {
             mn[j][i] = mn[j][i-] ;
             if(j+(<<(i-)) <= num)
  mn[j][i] = min( mn[j][i] , mn[j+(<<(i-))][i-]) ;
         }
     }
 }
 int rmq(int l,int r){
     int m = floor(log((double)(r-l+))/log(2.0)) ;
     int a = min(mn[l][m], mn[r-(<<m)+][m] ) ;
     return a ;
 }
 /*****************************树状数组*******************************/
 int Lowbit(int t){ //求最小幂2^k
     return t&(-t) ;
 }
 int Sum(int end) { //求前n项和
     int sum =  ;
     while(end > )
     {
         sum += c[end];
         end -= Lowbit(end);
     }
     return sum;
 }
 void plus(int pos , int n, int num){ //对某个元素进行加法操作
     while(pos <= n)
     {
         c[pos] += num;
         pos += Lowbit(pos);
     }
 }
 八、划分树
 int SU[];
 int U[][];
 int toL[][];
 int n,m,a,b,k,t;
 inline int cmp(const void *a,const void *b){return *((int *)a) - *((int *)b) ;}
 void buildtree(int l,int r,int d){
     int i ;
     if (l==r) return;
     int mid = (l+r)>> , nowl = l , nowr = mid+ , midtol =  ;
     for (i = mid ;i>=l  &&   SU[i] == SU[mid] ; i--) ++midtol ;
     for (i = l ; i <= r ; i++)
     {
         toL[d][i] = i==l ?  : toL[d][i-];
         if (U[d][i] < SU[mid])
         {
             U[d+][nowl++] = U[d][i];
             ++toL[d][i];
         }
         else if (U[d][i]==SU[mid] && midtol)
         {
             U[d+][nowl++] = U[d][i];
             ++toL[d][i];
             --midtol;
         }
         else U[d+][nowr++] = U[d][i];
     }
     buildtree(l,mid,d+);
     buildtree(mid+,r,d+);
 }
 int answer(int a,int b,int k){
     int l = ,r = n,d = ;
     int ls,rs,mid;
     while (a != b)
     {
         ls = a==l ?  : toL[d][a-];
         rs = toL[d][b];
         mid = (l+r)>>;
         if (k <= rs - ls)
         {
             a = l+ls;
             b = l+rs-;
             r = mid;
         }
         else        {
             a = mid++a-l-ls;
             b = mid++b-l-rs;
             k -= rs-ls;
             l = mid+;
         }
         ++d;
     }
     return U[d][a];
 }
 int solve(){
         for (i=;i<=n;i++)
         {
             scanf("%d",&t);
             SU[i]=U[][i]=t;
         }
         sort(SU+, SU++n) ;
         buildtree( , n ,  );
             printf("%d\n",answer(a, b, k ) );
     }