浅谈二维树状数组

①前置知识

静态二维前缀和:

①:预处理递推:f[ i ][ j ] = f[ i - 1 ][ j ] + f[ i ][ j -1 ]  - f[ i - 1][ j - 1] + val[ i ][ j ].

②:左上角( X₁ , Y₁ ),右下角( X₂ , Y₂),这一段的区间和:f[ X₂ ][ Y₂]  - f[ X₂ ][ Y₁ - 1] -f[ X₁ - 1][ Y₂ ] + f[ X₁ - 1][Y₁ - 1].

其实画一下图就很好理解了,具体详细教程从dalao的这篇blog: 传送门.

②二维树状数组

I.(单点修改,区间查询)

考虑一个点( X , Y )的存在,我们过这个点分别做X轴,Y轴的平行线,把这个点看做矩形的左上角,发现它只对它右下角的矩形才产生贡献.我们不由地联想到树状数组,用tree[ x ][ y ]的二维去维护点对'x下方', 'y右方'的贡献(想一想lowbit的作用).又因为树状数组求的和为前缀和,所以只要套静态二维前缀的的区间查询公式即可.

以LOJ的板子为例:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define e exit(0)
#define re register
#define LL long long
const int maxn = (1<<13);
LL n,m,flag,x1,y1,tree[maxn][maxn];
inline long long fd(){
    LL s=1,t=0;
    char  c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
            s=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        t=t*10+c-'0';
        c=getchar();
    }
    return s*t;
}
LL lowbit(LL x){
    return x&(-x);
}
void add(LL x,LL y,LL v){
    for(re LL i=x;i<=n;i+=lowbit(i))
        for(re LL j=y;j<=m;j+=lowbit(j))
            tree[i][j] += v;
}
LL getans(LL x,LL y){
    LL sum = 0;
    for(re LL i=x;i;i-=lowbit(i))
        for(re LL j=y;j;j-=lowbit(j))
            sum += tree[i][j];
    return sum;
}
int main()
{ 
    n = fd(),m = fd();
    while((scanf("%lld%lld%lld",&flag,&x1,&y1)) != EOF){
        if(flag == 1){
            LL v = fd();
            add(x1,y1,v); 
        } 
        else if(flag == 2){
            LL x2 = fd(),y2 = fd();
            LL ans = getans(x2,y2)-getans(x2,y1-1)-getans(x1-1,y2)+getans(x1-1,y1-1);
            printf("%lld
",ans); 
        }
    }
    return 0;
} 

II.(区间修改,区间查询)

①:一维树状数组的区间修改区间查询.

将单点修改,区间查询的差分技巧运用,容易发现 Σ(p , i =1)a[ i ] =Σ( p, i = 1)Σ(i , j = 1)cf[ j ].容易发现cf[ 1 ]用了p次,cf[ 2 ]用了p - 1次......那么上式可化简为Σ(p,i = 1)cf[ i ]*(p-i+1),将公式展开变成(p + 1)*Σ(p , i = 1)cf[ i ]  - Σ(p , i =1)cf[ i ] * i.用树状数组去维护cf[ i ]与cf[ i ]*i的值即可.以LOJ例题为例:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define e exit(0)
#define re register
#define LL long long
const int maxn = 1e6+10;
long long n,q,a[maxn],tree1[maxn],tree2[maxn];
inline LL fd(){
    LL s = 1,t = 0;
    char c = getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
            s=-1;
        c = getchar();
    }
    while(c>='0'&&c<='9'){
        t = t*10+c-'0';
        c = getchar();
    }
    return s*t;
}
LL lowbit(LL x){
    return x&(-x);
}
void add(LL x,LL v){
    for(re LL i=x;i<=n;i+=lowbit(i))
        tree1[i] += v,tree2[i] += v*x;
}
LL ask(LL x){
    LL s = 0;
    for(re LL i=x;i;i-=lowbit(i))
        s += (x+1)*tree1[i] - tree2[i];
    return s;
}
int main()
{
    n = fd(),q = fd();
    for(re LL i=1;i<=n;++i)
        a[i] = fd();
    for(re LL i=1;i<=n;++i)
        add(i,a[i]-a[i-1]);
    while(q--){
        LL flag = fd();
        if(flag == 1){
            LL l = fd(),r = fd(),v = fd();
            add(l,v),add(r+1,-v);
        }
        else if(flag == 2){
            LL l = fd(),r = fd();
            printf("%lld
",ask(r)-ask(l-1));
        }
    }
    return 0;
}

②:二维树状数组的区间修改与查询.

①:我们需要类比一维数组的区间修改与查询,这时我们要去定义一个二维的差分数组cf[ i ][ j ]表示val[ i ][ j ]与val[ i -1][ j ] + val[ i ][ j -1 ]  - val[ i -1][ j -1]的差.

那么以一个点(x , y)为右下角的矩阵内元素个数为Σ(x , i = 1)Σ(y, j = 1)Σ(i,k = 1)Σ(j,t = 1)cf[ k ][ t ].

②:将上式展开,Σ(x , i = 1)Σ(y, j = 1)cf[ i ][ j ]*(x - i + 1)*(y - j + 1),再展开,(x + 1)( y + 1)*Σ(x , i = 1)Σ(y, j = 1)cf[ i ][ j ] - (y+1)Σ(x , i = 1)Σ(y, j = 1)cf[ i ][ j ]*i

- (x + 1)Σ(x , i = 1)Σ(y, j = 1)cf[ i ][ j ]*j + Σ(x , i = 1)Σ(y, j = 1)cf[ i ][ j ]*i*j.用树状数组去维护cf[ i ][ j ],cf[ i ][ j ]*i,cf[ i ][ j ]*j,cf[ i ][ j ]*i*j,四个信息即可.

③:同时注意,修改时有别于一维,(x1 , y1)为左上角,(x2 , y2)为右下角的矩阵加v时,(x1,y1) + v,(x1,y1 + 1) - v,(x2 + 1,y1)-v,(x2 + 1,y2 + 1) + v.

以LOJ例题为例:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define e exit(0)
#define re register
#define LL long long
const int maxn = 3010;
LL n,m,flag,t1[maxn][maxn],t2[maxn][maxn],t3[maxn][maxn],t4[maxn][maxn];
inline LL fd(){
    LL s=1,t=0;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
            s=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        t=t*10+c-'0';
        c=getchar();
    }
    return s*t;
}
LL lowbit(LL x){
    return x&(-x);
}
void add(LL x,LL y,LL v){
    for(re LL i=x;i<=n;i+=lowbit(i))
        for(re LL j=y;j<=m;j+=lowbit(j)){
            t1[i][j] += v;
            t2[i][j] += v*x;
            t3[i][j] += v*y;
            t4[i][j] += v*x*y;
        }
}
void rang_add(LL x1,LL y1,LL x2,LL y2,LL v){
    add(x1,y1,v);
    add(x2+1,y1,-v);
    add(x1,y2+1,-v);
    add(x2+1,y2+1,v);
}
LL ask(LL x,LL y){
    LL s = 0;
    for(re LL i=x;i;i-=lowbit(i))
        for(re LL j=y;j;j-=lowbit(j))
            s += (x+1)*(y+1)*t1[i][j]-(y+1)*t2[i][j]-(x+1)*t3[i][j]+t4[i][j];
    return s;
}
LL rang_ask(LL x1,LL y1,LL x2,LL y2){
    return ask(x2,y2)-ask(x2,y1-1)-ask(x1-1,y2)+ask(x1-1,y1-1);
}
int main()
{
    n = fd(),m = fd();
    while(scanf("%lld",&flag)!=EOF){
        if(flag == 1){
            LL x1 = fd(),y1 = fd(),x2 = fd(),y2 = fd(),v = fd();
            rang_add(x1,y1,x2,y2,v);
        }
        else if(flag == 2){
            LL x1 = fd(),y1 = fd(),x2 = fd(),y2 = fd();
            LL ans = rang_ask(x1,y1,x2,y2);
            printf("%lld
",ans);
        }
    }
    return 0;
}

后记:考虑无修改操作时,n,m>=1e6,查询矩阵内元素个数,传送门.

附上dalao blog.