树状数组review

使用场景

有前缀思想，且需O(1e5)时间查询

例子

leetcode72双周T4

把数组1的地址映射到数组2上，对其进行前缀求和，即是前面出现了多少次，再求一次后缀，把结果累加

函数

插入

void inc(int x,int ad)
{
  for(; x<= n; x += x&-x) f[x]+=ad;
}

求和

int calc(int x){
  int Sum = 0;
  for(; x ; x-= x&-x ) Sum += f[x];
  return Sum;

例子的解

class Solution {
public:
    
    int f[100001], c[100001];
    long long v[100001];
    
    inline void inc(int x) {
        for (; x <= 100000; x += x & (-x))
            f[x]++;
    }
    
    int calc(int x) {
        int res = 0;
        for (; x; x -= x & (-x))
            res += f[x];
        return res;
    }
    
    long long goodTriplets(vector<int>& a, vector<int>& b) {
        long long ans = 0;
        int n = a.size();
        memset(f, 0, sizeof(f));
        for (int i = 0; i < n; i++)
            c[b[i]] = i;
        for (int i = 0; i < n; i++) {
            v[i] = calc(c[a[i]] + 1);
            inc(c[a[i]] + 1);
        }
        memset(f, 0, sizeof(f));
        for (int i = n - 1; i >= 0; i--) {
            v[i] *= calc(100000) - calc(c[a[i]] + 1);
            inc(c[a[i]] + 1);
        }
        for (int i = 0; i < n; i++)
            ans += v[i];
        return ans;
        
        
    }
};

板子

using LL = long long;
struct BIT{
    int n;
    vector<LL> v;
    BIT(int n) : n(n), v(n){
    }
    void add(int x, LL y) {
        for (x += 1; x <= n; x += x & -x) v[x - 1] += y;
    }
    LL sum(int x) {
        LL res = 0;
        for (x += 1; x; x -= x & -x) res += v[x - 1];
        return res;
    }
};
class Solution {
public:
    long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        vector<int> p(n), a(n);
        for (int i = 0; i < n; i += 1) p[nums1[i]] = i;
        for (int i = 0; i < n; i += 1) a[i] = p[nums2[i]];
        vector<LL> bit(n);
        BIT s(n), s2(n);
        LL ans = 0;
        for (int i = 0; i < n; i += 1) {
            ans += s2.sum(a[i]);
            s2.add(a[i], s.sum(a[i]));
            s.add(a[i], 1);
        }
        return ans;
    }
};