BZOJ1452 [JSOI2009]Count Solution

题意：自行脑补

做法：直接开权值那么多的二维树状数组暴力。

Code:

#include <cstdio>
#include <cctype>
#include <iostream>
#include <algorithm>
using namespace std;
 
inline int getc() {
    static const int L = 1 << 15;
    static char buf[L], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, L, stdin);
        if (S == T)
            return EOF;
    }
    return *S++;
}
inline int getint() {
    int c;
    while(!isdigit(c = getc()));
    int tmp = c - '0';
    while(isdigit(c = getc()))
        tmp = (tmp << 1) + (tmp << 3) + c - '0';
    return tmp;
}
 
int A[101][301][301], w[301][301];
int n, m;
void modify(int c, int x, int y, int add) {
    int t;
    for(; x <= n; x += x & -x) {
        for(t = y; t <= m; t += t & -t)
            A[c][x][t] += add;
    }
}
int get(int c, int x, int y) {
    int res = 0, t;
    for(; x; x -= x & -x) {
        for(t = y; t; t -= t & -t)
            res += A[c][x][t];
    }
    return res;
}
 
int main() {
    n = getint();
    m = getint();
     
    register int i, j;
    int x;
    for(i = 1; i <= n; ++i)
        for(j = 1; j <= m; ++j) {
            x = getint();
            w[i][j] = x;
            modify(x, i, j, 1);
        }
     
    int Q = getint();
    int ope, x1, y1, x2, y2, c;
    while(Q--) {
        ope = getint();
        if (ope == 1) {
            x1 = getint(), y1 = getint(), c = getint();
            modify(w[x1][y1], x1, y1, -1);
            modify(w[x1][y1] = c, x1, y1, 1);
        }
        else {
            x1 = getint(), x2 = getint(), y1 = getint(), y2 = getint();
            c = getint();
            printf("%d\n", get(c, x2, y2) - get(c, x1 - 1, y2) - get(c, x2, y1 - 1) + get(c, x1 - 1, y1 - 1));
        }
    }
     
    return 0;
}