【JSOI 2009】 Count

【题目链接】

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【算法】

            二维树状数组

【代码】

           

#include<bits/stdc++.h>
using namespace std;
#define MAXN 300
#define MAXC 100

int N,M,Q,i,j,opt,x,y,xa,ya,xb,yb,c;
int a[MAXN+10][MAXN+10];

template <typename T> inline void read(T &x) {
        int f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}

template <typename T> inline void write(T x) {
    if (x < 0) { putchar('-'); x = -x; }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}

template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

struct BinaryIndexedTree {
        int bit[MAXC+10][MAXN+10][MAXN+10];
        int lowbit(int x) { return x & (-x); }
        inline void modify(int c,int x,int y,int val) {
                int i,j;
                for (i = x; i <= N; i += lowbit(i)) {
                        for (j = y; j <= M; j += lowbit(j)) {
                                bit[c][i][j] += val;
                        }
                }
        }
        inline int query(int c,int x,int y) {
                int i,j,ret=0;
                for (i = x; i >= 1; i -= lowbit(i)) {
                        for (j = y; j >= 1; j -= lowbit(j)) {
                                ret += bit[c][i][j];
                        }
                }
                return ret;
        }
        inline int query(int c,int xa,int xb,int ya,int yb) {
                return query(c,xb,yb) - query(c,xb,ya-1) - query(c,xa-1,yb) + query(c,xa-1,ya-1);
        } 
} BIT;

int main() {
        
        read(N); read(M);
        for (i = 1; i <= N; i++) {
                for (j = 1; j <= M; j++) {
                        read(a[i][j]);
                        BIT.modify(a[i][j],i,j,1);                
                }
        }
        read(Q);
        while (Q--) {
                read(opt);
                if (opt == 1) {
                        read(x); read(y); read(c);
                        BIT.modify(a[x][y],x,y,-1);
                        BIT.modify(c,x,y,1);
                        a[x][y] = c;
                }    else {
                        read(xa); read(xb); read(ya); read(yb); read(c);
                        writeln(BIT.query(c,xa,xb,ya,yb));
                }
        }
        
        return 0;
    
}