3809: Gty的二逼妹子序列

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 ;
 }