「Bzoj 1452」「JSOI2009」Count (二维树状数组)

BZOJ 1452
可以每一种颜色建一个二维树状数组，数据$O(n^2c)$的空间复杂度可以接受

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ms(i,j) memset(i, j, sizeof i);
using namespace std;
int n,m,Q;
int a[300 + 5][300 + 5];
int lowbit(int x)
{
    return x&(-x);
}
struct bits
{
       int a[300 + 5][300 + 5];
       void init()
       {
            ms(a, 0);
       }
       void add(int x, int y, int c)
       {
            for (int i=x;i<=n;i+=lowbit(i))
            for (int j=y;j<=n;j+=lowbit(j))
            {
                a[i][j] += c;
            }
       }
       int sum(int x, int y)
       {
           int ret = 0;
           for (int i=x;i>0;i-=lowbit(i))
           for (int j=y;j>0;j-=lowbit(j))
           {
                ret += a[i][j];
           }
           return ret;
       }
       int query(int x1, int y1, int x2, int y2)
       {
           return sum(x2, y2)+sum(x1-1, y1-1)-sum(x1-1, y2)-sum(x2, y1-1);
       }
}b[100 + 5];
int main()
{
   // freopen("bzoj1452.in", "r", stdin); freopen("bzoj1452.out", "w", stdout);
    for (int i=1;i<=102;i++) b[i].init();
    scanf("%d%d", &n, &m);
    for (int i=1;i<=n;i++)
    for (int j=1;j<=m;j++)
    {
        scanf("%d", &a[i][j]);
        b[a[i][j]].add(i,j,1);
    }
    scanf("%d", &Q);
    for (int i=1;i<=Q;i++)
    {
        int opt;
        scanf("%d", &opt);
        if (opt==1)
        {
            int x, y, c;
            scanf("%d%d%d", &x, &y, &c);
            b[a[x][y]].add(x,y,-1);
            b[a[x][y]=c].add(x,y,1);
        } else
        {
            int x1, x2, y1, y2, c;
            scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &c);
            printf("%d\n", b[c].query(x1,y1,x2,y2));
        }
    }
    return 0;
}