问题抽象：区间内恰好出现K次的数的个数。 ------------------------------------------------------------------ UESTC出的题就是神啊T_T。。。一开始想了个函数式线段树方法后来发现错了=。=，然后也没什么思路，就是找着官方题解的方法做的。 思路： 题解说的用树状数组，这里当然也可以用线段树维护，和上面一样，线段树第j个数表示区间[j, i]内出现k次的数有多少个，然后像题解一样维护即可。（这种维护方法值得好好研究&&学习呀~） 代码中也有比较详细的注释：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)>>1)using namespace std;
const int maxn = 100100;int id,n,K;
int ans[maxn];
int w[maxn],wb[maxn];
int vis[maxn];
int a[maxn];            //线性权值
int l[maxn],r[maxn];    //线性区间
vector  v[maxn];   //边表
vector  pos[maxn]; //记录某数出现的位置
int num[maxn];          //记录某个数出现多少次了
map   M;      //离散化
int mtot;struct ANS
{int l,r;int id;
}Q[maxn];bool cmp(ANS a1, ANS a2)
{return a1.r ::iterator vp;if (v[x].size())for (vp = v[x].begin(); vp != v[x].end(); vp ++)dfs(*vp);r[x] = id;}
}int sum[maxn<<2],add[maxn<<2];
void build(int l,int r,int rt)
{sum[rt] = 0;add[rt] = 0;if (l == r)  return ;int mid = MID(l,r);build(l,mid,rt<<1);build(mid+1,r,rt<<1|1);
}
void pushdown(int rt,int w)
{if (add[rt]){add[rt<<1] += add[rt];add[rt<<1|1] += add[rt];sum[rt<<1] += add[rt] * (w - (w >> 1));sum[rt<<1|1] += add[rt] * (w >> 1);add[rt] = 0;}
}
void update(int s,int t,int v,int l,int r,int rt)
{if (s <= l && r <= t){sum[rt] += v * (r - l + 1);add[rt] += v;return ;}pushdown(rt, r-l+1);int mid = MID(l,r);if (s <= mid)   update(s,t,v,l,mid,rt<<1);if (mid < t)    update(s,t,v,mid+1,r,rt<<1|1);
}
int query(int p,int l,int r,int rt)
{if (l == p && r == p){return sum[rt];}pushdown(rt,r-l+1);int mid = MID(l,r);if (p <= mid)   return query(p,l,mid,rt<<1);else return query(p,mid+1,r,rt<<1|1);
}
int main()
{//freopen("data.txt","r+",stdin);int tt,caseo = 1;scanf("%d",&tt);while(tt--){//Initializemtot = id = 0;memset(v,0,sizeof(v));memset(vis,0,sizeof(vis));memset(pos,0,sizeof(pos));memset(num,0,sizeof(num));//inputprintf("Case #%d:\n",caseo ++);scanf("%d%d",&n,&K);for (int i = 1; i <= n; i ++)scanf("%d",&w[i]);for (int i = 1; i < n; i ++){int a,b;scanf("%d%d",&a,&b);v[a].push_back(b);  //边表}//树形结构转线性结构dfs(1);M.clear();//权值离散化for (int i = 1; i <= n; i ++)if (!M[a[i]])   M[a[i]] = ++ mtot;int q;scanf("%d",&q);for (int i = 0; i < q; i ++){int p;scanf("%d",&p);Q[i].l = l[p];Q[i].r = r[p];Q[i].id = i;}sort(Q,Q+q,cmp);int pt = 0;build(1,n,1);for (int i = 1; i <= n; i ++){pos[M[a[i]]].push_back(i);num[M[a[i]]] ++;if (num[M[a[i]]] >= K)if (num[M[a[i]]] == K)update(1,pos[M[a[i]]][0],1,1,n,1);else{int ss = (num[M[a[i]]] - K <= 2)?1:(num[M[a[i]]] - K - 2);update(ss,pos[M[a[i]]][num[M[a[i]]]-K-1],-1,1,n,1);update(pos[M[a[i]]][num[M[a[i]]]-K-1]+1,pos[M[a[i]]][num[M[a[i]]]-K],1,1,n,1);}else;while(pt < q && Q[pt].r == i){ans[Q[pt].id] = query(Q[pt].l,1,n,1);pt ++;}}for (int i = 0; i < q; i ++)printf("%d\n",ans[i]);if (tt) printf("\n");}return 0;
}