Educational Codeforces Round 22 E. Army Creation 主席树 或 分块

http://codeforces.com/contest/813/problem/E

题目大意：

给出长度为n的数组和k,  大小是1e5级别。

要求在线询问区间[l, r]权值,  权值定义为对于所有不同元素x在区间出现的次数和, 如果x出现次数>k, 那么按k算。

重要转换： 考虑一个区间[L, R]的某个数A[i], 它对答案有贡献 当且仅当 它前面与他权值相同的数中第k个数的位置(记为B[i]) < L

预处理B[], 每次询问就转化为 区间[L, R]中有多少个B[i] < L

可以用主席树 或者 分块解决。

也可以用此题的方法求区间不同元素个数, 其实就是k = 1的情况。

主席树代码：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <queue>
 using namespace std;
 
 typedef long long ll;
 
 #define N 100050
 const int INF =  << ;
 const double pi = acos(-);
 
 int pt;
 int a[N], b[N], root[N], lc[N * ], rc[N * ], cnt[N * ];
 vector<int> lis[N];
 
 int Add(int y, int l, int r, int v)
 {
     int x = ++pt;
     cnt[x] = cnt[y] + ;
     if (l < r)
     {
         int mid = (l + r) >> ;
         if (v <= mid)
         {
             rc[x] = rc[y];
             lc[x] = Add(lc[y], l, mid, v);
         }
         else
         {
             lc[x] = lc[y];
             rc[x] = Add(rc[y], mid + , r, v);
         }
     }
     return x;
 }
 
 int Query(int x, int y, int L, int R, int l, int r)
 {
     //cout << L <<" " << R << " " << cnt[x] << " " << cnt[y] << endl;
     if (l > R || r < L) return ;
     if (l <= L && r >= R) return cnt[x] - cnt[y];
 
     int Mid = (L + R) >> ;
     return Query(lc[x], lc[y], L, Mid, l, r) + Query(rc[x], rc[y], Mid + , R, l, r);
 }
 
 int main()
 {
     //freopen("in.in", "r", stdin);
     //freopen("out.out", "w", stdout);
 
      int n, k, Q, lastans = , l, r;
     scanf("%d %d", &n, &k);
     for (int i = ; i <= n; ++i) scanf("%d", &a[i]), lis[a[i]].push_back(i);
 
     for (int i = ; i <= ; ++i)
     {
         if (lis[i].size() <= k) continue;
         for (int j = k; j < lis[i].size(); ++j) b[lis[i][j]] = lis[i][j - k];
     }
 
     root[] = ++pt;
     for (int i = ; i <= n; ++i) root[i] = Add(root[i - ], , n, b[i]);
 
     scanf("%d", &Q);
     while (Q--)
     {
         scanf("%d %d", &l, &r);
         l = (l + lastans) % n + ;
         r = (r + lastans) % n + ;
         if (l > r) swap(l, r);
         lastans = Query(root[r], root[l - ], , n, , l - );
         printf("%d\n", lastans);
     }
 
     return ;
 } 

分块代码：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <queue>
 using namespace std;
 
 typedef long long ll;
 
 #define N 100050
 const int INF =  << ;
 const double pi = acos(-);
 
 int pt;
 int a[N], b[N], id[N], dp[][N];
 int bl[], br[];
 vector<int> lis[N];
 
 int main()
 {
     //freopen("in.in", "r", stdin);
     //freopen("out.out", "w", stdout);
 
      int n, k, Q, lastans = , l, r;
     scanf("%d %d", &n, &k);
     for (int i = ; i <= n; ++i) scanf("%d", &a[i]), lis[a[i]].push_back(i);
 
     for (int i = ; i <= ; ++i)
     {
         if (lis[i].size() <= k) continue;
         for (int j = k; j < lis[i].size(); ++j) b[lis[i][j]] = lis[i][j - k];
     }
 
     int block = (int)sqrt(n + );
 
     for (int i = ; ; ++i)
     {
         l = (i - ) * block + ;
         r = min(n, l + block - );
         bl[i] = l, br[i] = r;
         for (int j = l; j <= r; ++j) dp[i][b[j]]++, id[j] = i;
         for (int j = ; j <= ; ++j) dp[i][j] += dp[i][j - ];
         if (r == n) break;
     }
 
      scanf("%d", &Q);
      while (Q--)
      {
          scanf("%d %d", &l, &r);
           l = (l + lastans) % n + ;
           r = (r + lastans) % n + ;
           if (l > r) swap(l, r);
 
           int res = ;
           if (id[l] == id[r])
           {
               for (int i = l; i <= r; ++i)
                   res += b[i] < l;
         }
         else
         {
             for (int i = l; i <= br[id[l]]; ++i) res += b[i] < l;
             for (int i = bl[id[r]]; i <= r; ++i) res += b[i] < l;
             for (int i = id[l] + ; i <= id[r] - ; ++i) res += dp[i][l - ];
         }
         printf("%d\n", res);
         lastans = res;
     }
 
     return ;
 }