bzoj 1452: [JSOI2009]Count ——二维树状数组

escription

Input

Output

Sample Input

Sample Output

1
2

HINT

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这道题是裸的二维树状数组.....直接每个颜色弄一个二维的树状数组然后容斥（也不知道算不算）就可以辣

#include<cstdio>
#include<cstring>
#include<algorithm>
int read(){
    int ans=,f=,c=getchar();
    while(c<''||c>''){if(c=='-') f=-; c=getchar();}
    while(c>=''&&c<=''){ans=ans*+(c-''); c=getchar();}
    return ans*f;
}
int n,m,q,k,f[][];
int s[][][];
#define lowbit(x) x&-x
void ins(int b[][],int x,int y,int v){
    for(int i=x;i<=n;i+=lowbit(i))
    for(int j=y;j<=m;j+=lowbit(j)) b[i][j]+=v;
}
int query(int b[][],int x,int y){
    int sum=;
    for(int i=x;i;i-=lowbit(i))
    for(int j=y;j;j-=lowbit(j)) sum+=b[i][j];
    return sum;
}
int main(){
    int x1,y1,x2,y2,c;
    n=read(); m=read();
    for(int i=;i<=n;i++)
    for(int j=;j<=m;j++)
    k=read(),ins(s[k],i,j,),f[i][j]=k;
    q=read();
    for(int i=;i<=q;i++){
        k=read();
        if(k==){
            x1=read(); y1=read(); c=read();
            ins(s[f[x1][y1]],x1,y1,-);
            f[x1][y1]=c;
            ins(s[f[x1][y1]],x1,y1,);
        }
        else{
            x1=read(); x2=read();
            y1=read(); y2=read();
            c=read();
            printf("%d\n",query(s[c],x2,y2)-query(s[c],x2,y1-)-query(s[c],x1-,y2)+query(s[c],x1-,y1-));
        }
    }
    return ;
}