AcWing算法提高课 树状数组

解决的问题：

1、快速求前缀和（logn）（区间求和）

2、修改某一个数（logn）（单点修改）

对比：数组：求前缀和On，修改O1；前缀和数组：求前缀和O1，修改On

 

 

 

 C[x]表示，以x结尾，长度为lowbit(x)的区间中的计数

 

 

 

 

 查询：

修改：

 

 

模板：

int a[200010];
LL tr[200010];
int n;

int lowbit(int x)
{
    return x & (-x);
}
int add(int x, int c)
{
    for (int i = x; i <= n; i += lowbit(i))
    {
        tr[i] += c;
    }
    return 0;

}
LL sum(int x)
{
    LL res = 0;
    for (int i = x; i > 0; i -= lowbit(i))
        res += tr[i];
    return res;

}

 

 

 对称问题：

区间修改，单点查询

可以转化为对差分数组的单点（两点）修改，区间查询

例题：

AcWing 242. 一个简单的整数问题 

#include<bits/stdc++.h>
#include<unordered_set>
using namespace std;
typedef long long LL;

LL a[100010];
LL diff[100010];
LL tr[100010];
int n, m;
LL lowbit(int x)
{
    return x & (-x);
}
void add(LL x, LL c)
{
    for (int i = x; i <= n; i += lowbit(i))
    {
        tr[i] += c;
    }
}
LL sum(LL x)
{
    LL res = 0;
    for (int i = x; i > 0; i -= lowbit(i))
    {
        res += tr[i];
    }
    return res;
}
void YD()
{

    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i], diff[i] = a[i] - a[i - 1];
    for (int i = 1; i <= n; i++)
    {
        add(i, diff[i]);
    }
    for (int i = 1; i <= m; i++)
    {
        char op; cin >> op;
        if (op == 'Q')
        {
            int x; cin >> x;
            cout << sum(x) << endl;
        }
        else
        {
            LL l, r, d;
            cin >> l >> r >> d;
            add(r+1, -d);
            add(l, d);
        }
    }

}

int main()
{
    YD();
    return 0;
}

 拓展问题：

区间修改，区间查询

修改依旧使用差分数组。

区间查询，前缀和的前缀和，转化为不同形式的前缀和的差

 

例题：AcWing 243. 一个简单的整数问题2

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

LL a[100010];
int n, m;
LL tr1[100010];
LL tr2[100010];
LL lowbit(LL x)
{
    return x & (-x);
}
void add(LL x, LL c)
{
    for (int i = x; i <= n; i += lowbit(i))
    {
        tr1[i] += c;
    }
}
LL sum(LL x)
{
    LL res = 0;
    for (int i = x; i >0; i -= lowbit(i))
    {
        res += tr1[i];
    }
    return res;
}
void add2(LL x, LL c)
{
    for (int i = x; i <= n; i += lowbit(i))
    {
        tr2[i] += c;
    }
}
LL sum2(LL x)
{
    LL res = 0;
    for (int i = x; i > 0; i -= lowbit(i))
    {
        res += tr2[i];
    }
    return res;
}
void YD()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
    {
        add(i, a[i] - a[i - 1]);
        add2(i, (a[i] - a[i - 1])*i);
    }
    while (m--)
    {
        char op; cin >> op;
        if (op == 'C')
        {
            LL l, r, d;
            cin >> l >> r >> d;
            add(l, d);
            add(r + 1, -d);
            add2(l, l*d);
            add2(r + 1, -d*(r+1));
        }
        else
        {
            LL l, r;
            cin >> l >> r ;
            cout << sum(r)*(r + 1) - sum2(r)
                - (sum(l - 1)*(l)-sum2(l - 1))
                << endl;


        }
    }
}

int main()
{
    YD();
return 0;
}

树状数组结合二分例题

244. 谜一样的牛

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

int a[100010];
int n, m;
int h[100010];
int tr[100010];
int lowbit(int i)
{
    return i & (-i);
}
void add(int x, int c)
{
    for (int i = x; i <= n; i += lowbit(i))
        tr[i] += c;
}
int sum(int x)
{
    int res = 0;
    for (int i = x; i > 0; i -= lowbit(i))
        res += tr[i];
    return res;
}
void YD()
{
    cin >> n;
    a[1] = 0;
    for (int i = 2; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++)
    {
        tr[i] = lowbit(i);
    }
    for (int i = n; i >= 0; i--)
    {
        int index = a[i] + 1;
        int l = 1, r = n;
        while (l < r)
        {
            int mid = (l + r) / 2;
            if (sum(mid) >= index)
                r = mid;
            else
                l = mid + 1;
        }
        h[i] = l;
        add(l, -1);
    }
    for (int i = 1; i <= n; i++) cout << h[i] << endl;
    
}

int main()
{
    YD();
return 0;
}