A one-dimensional array of two-dimensional template tree

Fenwick tree can also be said to be binary index tree, it is an advanced data structures, but in fact is easy to use.

Reference herein blog as follows:

http://www.cppblog.com/menjitianya/archive/2015/11/02/212171.html

https://blog.csdn.net/zars19/article/details/54620021

https://www.cnblogs.com/RabbitHu/p/BIT.html

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 1, single-point range query modification +

int C [MAXN];
 int lowbit ( int X) // Get the least significant bit. 1 
{
     return X & (- X); 
} 
void Update ( int X, int C) // single point updating 
{
     for ( int I = X; I <= n-; I + = lowbit (I)) 
    { 
        C [I] + = C; 
     } 
} 
int query ( int X) // query prefix and C [. 1 ... X] 
{
     int ANS = 0 ;
     for ( int i = x; i> =0 ; I- = lowbit (I)) 
    { 
        ANS + = C [I]; 
    } 
    return ANS; 
} 
int rgquery ( int L, int R & lt) // Interval Query 
{
     return Query (R & lt) -query (L- . 1 ) ; 
 }

View Code

2, modify the interval + single point of inquiry

 

 1 int c1[maxn];//拆分数列 
 2 void updata(int x,int y) //a[x-n]区间修改 
 3 {
 4     for(int i=x;i<=n;i+=lowbit(i))
 5     {
 6         c1[i]+=y;
 7     }
 8 }
 9 void range_updata(int l,int r,int val)
10 {
11     updata(l,val);
12     updata(r+1,-val);
13 }
14 
15 int query(int x)
16 {
17     int ans=0;
18     for(int i=x;i;i-=lowbit(i))
19     {
20         ans+=c1[i];
21     }
22     return ans;
23 }

View Code

 

3, the section modify the query interval +

 

 

 

 

 

 1 void add(ll p, ll x){
 2     for(int i = p; i <= n; i += i & -i)
 3         sum1[i] += x, sum2[i] += x * p;
 4 }
 5 void range_add(ll l, ll r, ll x){
 6     add(l, x), add(r + 1, -x);
 7 }
 8 ll ask(ll p){
 9     ll res = 0;
10     for(int i = p; i; i -= i & -i)
11         res += (p + 1) * sum1[i] - sum2[i];
12     return res;
13 }
14 ll range_ask(ll l, ll r){
15     return ask(r) - ask(l - 1);
16 }

 

二维树状数组

在一维树状数组中，tree[x]（树状数组中的那个“数组”）记录的是右端点为x、长度为lowbit(x)的区间的区间和。
那么在二维树状数组中，可以类似地定义tree[x][y]记录的是右下角为(x, y)，高为lowbit(x), 宽为 lowbit(y)的区间的区间和。

1、单点修改+区间查询

void add(int x, int y, int z){ //将点(x, y)加上z
    int memo_y = y;
    while(x <= n){
        y = memo_y;
        while(y <= n)
            tree[x][y] += z, y += y & -y;
        x += x & -x;
    }
}
void ask(int x, int y){//求左上角为(1,1)右下角为(x,y) 的矩阵和
    int res = 0, memo_y = y;
    while(x){
        y = memo_y;
        while(y)
            res += tree[x][y], y -= y & -y;
        x -= x & -x;
    }
}

 

2、区间修改+单点查询

 

 

 

 

 

void add(int x, int y, int z){ 
    int memo_y = y;
    while(x <= n){
        y = memo_y;
        while(y <= n)
            tree[x][y] += z, y += y & -y;
        x += x & -x;
    }
}
void range_add(int xa, int ya, int xb, int yb, int z){
    add(xa, ya, z);
    add(xa, yb + 1, -z);
    add(xb + 1, ya, -z);
    add(xb + 1, yb + 1, z);
}
void ask(int x, int y){
    int res = 0, memo_y = y;
    while(x){
        y = memo_y;
        while(y)
            res += tree[x][y], y -= y & -y;
        x -= x & -x;
    }
}

 

3、区间修改+区间查询

 

 

 

 

 

 

 1 const int N = 205;
 2 ll n, m, Q;
 3 ll t1[N][N], t2[N][N], t3[N][N], t4[N][N];
 4 void add(ll x, ll y, ll z){
 5     for(int X = x; X <= n; X += X & -X)
 6         for(int Y = y; Y <= m; Y += Y & -Y){
 7             t1[X][Y] += z;
 8             t2[X][Y] += z * x;
 9             t3[X][Y] += z * y;
10             t4[X][Y] += z * x * y;
11         }
12 }
13 void range_add(ll xa, ll ya, ll xb, ll yb, ll z){ //(xa, ya) 到 (xb, yb) 的矩形
14     add(xa, ya, z);
15     add(xa, yb + 1, -z);
16     add(xb + 1, ya, -z);
17     add(xb + 1, yb + 1, z);
18 }
19 ll ask(ll x, ll y){
20     ll res = 0;
21     for(int i = x; i; i -= i & -i)
22         for(int j = y; j; j -= j & -j)
23             res += (x + 1) * (y + 1) * t1[i][j]
24                 - (y + 1) * t2[i][j]
25                 - (x + 1) * t3[i][j]
26                 + t4[i][j];
27     return res;
28 }
29 ll range_ask(ll xa, ll ya, ll xb, ll yb){
30     return ask(xb, yb) - ask(xb, ya - 1) - ask(xa - 1, yb) + ask(xa - 1, ya - 1);
31 }