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 = 1 << 30;
const double pi = acos(-1);

int pt;
int a[N], b[N], root[N], lc[N * 60], rc[N * 60], cnt[N * 60];
vector<int> lis[N];

int Add(int y, int l, int r, int v)
{
    int x = ++pt;
    cnt[x] = cnt[y] + 1;
    if (l < r)
    {
        int mid = (l + r) >> 1;
        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 + 1, 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 0;
    if (l <= L && r >= R) return cnt[x] - cnt[y];
    
    int Mid = (L + R) >> 1;
    return Query(lc[x], lc[y], L, Mid, l, r) + Query(rc[x], rc[y], Mid + 1, R, l, r);
}

int main()
{
    //freopen("in.in", "r", stdin);
    //freopen("out.out", "w", stdout);
 
     int n, k, Q, lastans = 0, l, r;
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), lis[a[i]].push_back(i);
    
    for (int i = 1; i <= 100000; ++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[0] = ++pt;
    for (int i = 1; i <= n; ++i) root[i] = Add(root[i - 1], 0, n, b[i]);
    

    scanf("%d", &Q);
    while (Q--)
    {
        scanf("%d %d", &l, &r);
        l = (l + lastans) % n + 1;
        r = (r + lastans) % n + 1;
        if (l > r) swap(l, r);
        lastans = Query(root[r], root[l - 1], 0, n, 0, l - 1);
        printf("%d\n", lastans);
    }
     
    return 0;
} 

 

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 = 1 << 30;
const double pi = acos(-1);

int pt;
int a[N], b[N], id[N], dp[350][N];
int bl[350], br[350];
vector<int> lis[N];

int main()
{
    //freopen("in.in", "r", stdin);
    //freopen("out.out", "w", stdout);
 
     int n, k, Q, lastans = 0, l, r;
    scanf("%d %d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), lis[a[i]].push_back(i);
    
    for (int i = 1; i <= 100000; ++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 + 1);
    
    for (int i = 1; ; ++i)
    {
        l = (i - 1) * block + 1;
        r = min(n, l + block - 1);
        bl[i] = l, br[i] = r;
        for (int j = l; j <= r; ++j) dp[i][b[j]]++, id[j] = i;
        for (int j = 1; j <= 100000; ++j) dp[i][j] += dp[i][j - 1];
        if (r == n) break;
    }
     
     scanf("%d", &Q);
     while (Q--)
     {
         scanf("%d %d", &l, &r);
          l = (l + lastans) % n + 1;
          r = (r + lastans) % n + 1;
          if (l > r) swap(l, r);

          int res = 0;
          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] + 1; i <= id[r] - 1; ++i) res += dp[i][l - 1];
        }
        printf("%d\n", res);
        lastans = res;
    }
     
    return 0;
}