BZOJ 1452 Count(二维树状数组)

大水题。 建立100个二维树状数组,总复杂度就是O(qlognlogm).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=305;
//Code begin...

int tree[101][N][N], a[N][N], n, m;

void add(int x, int i, int j, int val){
    while (i<=n) {
        int k=j;
        while (k<=m) tree[x][i][k]+=val, k+=lowbit(k);
        i+=lowbit(i);
    }
}
int query(int x, int i, int j){
    int res=0;
    while (i) {
        int k=j;
        while (k) res+=tree[x][i][k], k-=lowbit(k);
        i-=lowbit(i);
    }
    return res;
}
int main ()
{
    int Q, flag, x1, y1, x2, y2, c;
    n=Scan(); m=Scan();
    FOR(i,1,n) FOR(j,1,m) {
        a[i][j]=Scan();
        add(a[i][j],i,j,1);
    }
    Q=Scan();
    while (Q--) {
        flag=Scan();
        if (flag==1) {
            x1=Scan(); y1=Scan(); c=Scan();
            add(a[x1][y1],x1,y1,-1); add(c,x1,y1,1);
            a[x1][y1]=c;
        }
        else {
            x1=Scan(); x2=Scan(); y1=Scan(); y2=Scan(); c=Scan();
            printf("%d\n",query(c,x2,y2)-query(c,x2,y1-1)-query(c,x1-1,y2)+query(c,x1-1,y1-1));
        }
    }
    return 0;
}