3809: Gty的二逼妹子序列

链接

分析：

　　和这道AHOI2013 作业差不多。权值是1~n的，所以对权值进行分块。$O(1)$修改，$O(\sqrt n)$查询。

代码：

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<iostream>

 #include<cmath>

 #include<cctype>

 #include<set>

 #include<queue>

 #include<vector>

 #include<map>

 using namespace std;

 typedef long long LL;

 inline int read() {

     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;

     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;

 }

 const int N = ;

 int bel[N], sum[N], a[N], Ans[N * ], cnt[N], B;

 struct Que{

     int l, r, a, b, id;

     bool operator < (const Que &A) const {

         return bel[l] == bel[A.l] ? r < A.r : bel[l] < bel[A.l];

     }

 }Q[N * ];

 inline void add(int x) {

     cnt[x] ++;

     if (cnt[x] == ) sum[bel[x]] ++;

 }

 inline void del(int x) {

     cnt[x] --;

     if (cnt[x] == ) sum[bel[x]] --;

 }

 inline int query(int l,int r) {

     int res = ;

     for (int i = l, lim = min(r, bel[l] * B); i <= lim; ++i) res += (cnt[i] > );

     if (bel[l] != bel[r])

         for (int i = (bel[r] - ) * B + ; i <= r; ++i) res += (cnt[i] > );

     for (int i = bel[l] + ; i <= bel[r] - ; ++i) res += sum[i];

     return res;

 }

 int main() {

     int n = read(), m = read(); B = sqrt(n);

     for (int i = ; i <= n; ++i) a[i] = read();

     for (int i = ; i <= m; ++i)

         Q[i].l = read(), Q[i].r = read(), Q[i].a = read(), Q[i].b = read(), Q[i].id = i;

     for (int i = ; i <= n; ++i) bel[i] = (i - ) / B + ;

     sort(Q + , Q + m + );

     int L = , R = ;

     for (int i = ; i <= m; ++i) {

         while (L > Q[i].l) add(a[--L]);

         while (R < Q[i].r) add(a[++R]);

         while (L < Q[i].l) del(a[L++]);

         while (R > Q[i].r) del(a[R--]);

         Ans[Q[i].id] = query(Q[i].a, Q[i].b);

     }

     for (int i = ; i <= m; ++i) printf("%d\n",Ans[i]);

     return ;

 }