bzoj1452 [JSOI2009]Count

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Description

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Input

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Output

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Sample Input

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Sample Output

1
2

HINT

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Solution

二维树状数组傻逼题，这里写一下，~~方便以后抄板，~~加深理解。

#include<bits/stdc++.h>using namespace std;inline int read() {    int x = 0, flag = 1; char ch = getchar();    while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }    while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }    return x * flag;}inline void write(int x) { if (x >= 10) write(x / 10); putchar(x % 10 + '0'); }#define N 305#define rep(i, a, b) for (int i = a; i <= b; i++)int n, m;int bit[N / 3][N][N];int g[N][N];inline int lowbit(int k) { return (k & (-k)); }void add(int x, int y, int num, int v) {    for (int i = x; i <= n; i += lowbit(i)) for (int j = y; j <= m; j += lowbit(j))        bit[num][i][j] += v;}int ask(int x, int y, int num) {    int ans = 0;    for (int i = x; i; i -= lowbit(i)) for (int j = y; j; j -= lowbit(j)) ans += bit[num][i][j];    return ans;}int main() {    cin >> n >> m;    rep(i, 1, n) rep(j, 1, m) g[i][j] = read(), add(i, j, g[i][j], 1);    int q = read();    while (q--) {        int t = read();        if (t == 1) {            int x = read(), y = read(), c = read();            add(x, y, g[x][y], -1); g[x][y] = c; add(x, y, c, 1);        }        else {            int x1 = read(), x2 = read(), y1 = read(), y2 = read(), c = read();            int ans = ask(x2, y2, c) - ask(x1 - 1, y2, c) - ask(x2, y1 - 1, c)                + ask(x1 - 1, y1 - 1, c);            write(ans), puts("");        }    }    return 0;}

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