hdu 4456 Crowd(二维树状数组)

题目链接：hdu 4456 Crowd

题目大意：给定N，然后M次操作

• 1 x y z：在x，y的位置加z
• 2 x y z：询问与x，y曼哈顿距离小于z的点值和。

解题思路：将矩阵旋转45度，然后询问就等于是询问一个矩形，可以用容斥定理搞，维护用二维树状数组，但是空间开

不下，直接用离散化，将有用到的点处理出来。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 4000005;
const int maxm = 80005;

#define lowbit(x) ((x)&(-x))
int N, M, W, E, H[maxn+5], fenw[maxn + 5];
int O[maxm], X[maxm], Y[maxm], Z[maxm];

inline int find (int x) {
    return lower_bound(H + 1, H + E, x) - H;
}

void hashPoint (int x, int y) {
    for (int i = x; i <= W; i += lowbit(i)) {
        for (int j = y; j <= W; j += lowbit(j))
            H[E++] = i * W + j;
    }
}

void add(int x, int y, int d) {
    for (int i = x; i <= W; i += lowbit(i)) {
        for (int j = y; j <= W; j += lowbit(j)) {
            int pos = find(i * W + j);
            fenw[pos] += d;
        }
    }
}

int sum (int x, int y) {
    int ret = 0;
    for (int i = x; i; i -= lowbit(i)) {
        for (int j = y; j; j -= lowbit(j)) {
            int pos = find(i * W + j);
            if (H[pos] == i * W + j)
                ret += fenw[pos];
        }
    }
    return ret;
}

void init () {
    E = 1;
    W = 2 * N;
    scanf("%d", &M);
    memset(fenw, 0, sizeof(fenw));

    for (int i = 1; i <= M; i++) {
        scanf("%d%d%d%d", &O[i], &X[i], &Y[i], &Z[i]);
        int x = X[i] - Y[i] + N;
        int y = X[i] + Y[i];
        if (O[i] == 1)
            hashPoint(x, y);
    }
    sort(H + 1, H + E);
    E = unique(H + 1, H + E) - H;
}

void solve() {
    for (int i = 1; i <= M; i++) {
        int x = X[i] - Y[i] + N;
        int y = X[i] + Y[i];

        if (O[i] == 1)
            add(x, y, Z[i]);
        else {
            int a = max(1, x - Z[i]);
            int b = max(1, y - Z[i]);
            int c = min(W, x + Z[i]);
            int d = min(W, y + Z[i]);
            printf("%d\n", sum(c, d) - sum(c, b-1) - sum(a-1, d) + sum(a-1, b-1));
        }
    }
}

int main () {
    while (scanf("%d", &N) == 1 && N) {
        init();
        solve();
    }
    return 0;
}