bzoj3211 花神游历各国

bzoj3211 花神游历各国

给一个序列，支持以下两种操作：

1. 区间开方
2. 区间和

(nleq10^5, mleq2 imes10^5)

线段树/树状数组+并查集

---

势能线段树，，不多说了

主要记录一下第二种做法的写法

时间复杂度 (O(mlog n))

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 1e5 + 10;
int n, m, a[maxn], par[maxn]; ll c[maxn];

int find(int x) {
  return par[x] == x ? x : par[x] = find(par[x]);
}

void add(int pos, int x) {
  for (; pos <= n; pos += pos & -pos) {
    c[pos] += x;
  }
}

ll query(int pos) {
  ll res = 0;
  for (; pos; pos &= pos - 1) {
    res += c[pos];
  }
  return res;
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    add(i, a[i]), par[i] = i;
  }
  par[n + 1] = n + 1;
  scanf("%d", &m);
  while (m--) {
    int op, l, r;
    scanf("%d %d %d", &op, &l, &r);
    if (op == 1) {
      printf("%lld
", query(r) - query(l - 1));
    } else {
      for (int i = find(l); i <= r; i = find(i + 1)) {
        int t = sqrt(a[i]);
        add(i, t - a[i]), a[i] = t;
        if (a[i] < 2) par[i] = find(i + 1);
      }
    }
  }
  return 0;
}