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zc
2019-12-22 14:39:00
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此题卡常,需要使用二维树状数组

联想一维的树状数组:

差分数组:d_0=a_0,d_i=a_i-a_{i-1}(i>0)

令差分数组的前缀和sd_i=\sum_{j=0}^id_j,那么sd_i=a_i

我们现在来看看二维前缀和

s(i,j)=s(i-1,j)+s(i,j-1)-s(i-1,j-1)+a(i,j)

查询(x_1,y_1)(x_2,y_2)的和则为:

s(x_2,y_2)-s(x_1-1,y_2)-s(x_2,y_1-1)+s(x_1-1,y_1-1)

我们可以使用差分数组:

d(i,j)=a(i,j)-(a(i-1,j)+a(i,j-1)-a(i-1,j-1))

修改

d(x_1,y_1)+=v

d(x_1,y_2+1)-=v

d(x_2+1,y_1)-=v

d(x_2+1,y_2+1)+=v

举个例子:

0 0 0 0 0
0 1 1 1 0
0 1 1 1 0 
0 0 0 0 0

1部分加上v,那么

0  0 0 0 0
0 +v 0 0 -v
0  0 0 0 0 
0 -v 0 0 +v

查询

对于点(x,y)它的值应该是:\sum\limits_{i=1}^x\sum\limits_{j=1}^y d(i,j)

那么(1,1)(x,y)的和就是:

$$

\sum{i=1}^x\sum{j=1}^y\sum{p=1}^i\sum{q=1}^j d(p,q) \ =\sum{i=1}^x\sum{j=1}^yd(i,j)\times(x-i+1)\times(y-j+1) \ =\sum{i=1}^x\sum{j=1}^yd(i,j)\times[(xy-xj+x)+(-yi+ij-i)+(y-j+1)] \ =\sum{i=1}^x\sum{j=1}^yd(i,j)\times (xy+x+y+1)-d(i,j)\times i(y+1)-d(i,j)\times j(x+1) + d(i,j)\times i \times j $$

我们需要维护d(i,j),d(i,j)\times i,d(i,j) \times j, d(i,j)\times i \times j

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 2049
#define lb(x) (x&-x)
int n,m;
struct bit{
    int s[N][N];
    void add(int x,int y,int v){
        for(int i=x;i<=n;i+=lb(i))
            for(int j=y;j<=m;j+=lb(j))
                s[i][j]+=v;
    }
    int ask(int x,int y){
        int ans=0;
        for(int i=x;i;i-=lb(i))
            for(int j=y;j;j-=lb(j))
                ans+=s[i][j];
        return ans;
    }
}d,di,dj,dij;
void upd(int x,int y,int v){
    d.add(x,y,v);
    di.add(x,y,v*x);
    dj.add(x,y,v*y);
    dij.add(x,y,v*x*y);
}
int ask(int x,int y){
    return
    d.ask(x,y)*(x*y+x+y+1)
    -di.ask(x,y)*(y+1)
    -dj.ask(x,y)*(x+1)
    +dij.ask(x,y);
}
int main(){
    in(n,m);
    char opt;
    int x,y,X,Y,v;
    while(!__){
        in(opt,x,y,X,Y);
        if(__)break;
        if(opt=='L')
            in(v),
            upd(x,y,v),
            upd(x,Y+1,-v),
            upd(X+1,y,-v),
            upd(X+1,Y+1,v);
        else out(ask(X,Y)-ask(x-1,Y)-ask(X,y-1)+ask(x-1,y-1),ln);
    }
    flush();
}
LG 4514 上帝造题的七分钟
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